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NCERT Solutions For Class 12 Chemistry Chapter 12 Aldehydes Ketones and Carboxylic Acids

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NCERT Solutions For Class 12 Chemistry Chapter 12 Aldehydes Ketones and Carboxylic Acids

NCERT INTEXT QUESTION

12.1 Write the structures of the following compounds:
(i)α-Methoxypropionaldehyde
(ii)3-Hydroxybutanal
(iii)2-Hydroxycyclopentane carbaldehyde
(iv)4-OxopentanaI
(v)Di-sec.butylketone
(vi)4-fluoroaeetophenone
Ans.
NCERT Solutions For Class 12 Chemistry Chapter 12 Aldehydes Ketones and Carboxylic Acids-1
NCERT Solutions For Class 12 Chemistry Chapter 12 Aldehydes Ketones and Carboxylic Acids-2

12.2 Write the structures of products of following reactions:
NCERT Solutions For Class 12 Chemistry Chapter 12 Aldehydes Ketones and Carboxylic Acids-3
Ans.
NCERT Solutions For Class 12 Chemistry Chapter 12 Aldehydes Ketones and Carboxylic Acids-4

12.3 Arrange the following compounds in increasing order of their boiling points:
CH3CHO, CH3CH2OH ,CH3OCH3, CH3CH2CH3
Ans.The order is : CH3CH2CH3 < CH3OCH3 < CH3CHO <CH3CH2OH
All these compounds have comparable molecular masses CH3CH2OH undergoes extensive intermolecular Il-bonding and thus its b.pt. is the highest. CH3CHO is more pdlar than CH3OCH3 so that dipole-dipoie interactions in CH3CHO are greater than in CH3OCH3. Thus, b.pt. of CH3CHO > CH3OCH3. CH3CH2CH3 has only weak van der waals forces between its molecules and hence has the lowest b.pt.

12.4 Arrange the following compounds in increasing order of their reactivity in nucleophilic addition reactions
(i)Ehtanal, propanaL, propanone, butanone
(ii)Benzaldehyde,p-Tolualdehyde, p-Nitrobenzaldehyde, acetophenone.
Ans. (i) Butanone < Propanone < Propanal < Ethanal .This is because as the no. of alkyl groups attached to carbonyl carbon increases, +I-effect increases. As a result, e density
NCERT Solutions For Class 12 Chemistry Chapter 12 Aldehydes Ketones and Carboxylic Acids-5

12.5.Predict the products of the following reactions:
NCERT Solutions For Class 12 Chemistry Chapter 12 Aldehydes Ketones and Carboxylic Acids-6
NCERT Solutions For Class 12 Chemistry Chapter 12 Aldehydes Ketones and Carboxylic Acids-7
Ans.
NCERT Solutions For Class 12 Chemistry Chapter 12 Aldehydes Ketones and Carboxylic Acids-8

12.6 Give the 1UPAC names of the following compounds:
(i)PhCH2CH2COOH
(ii)(CH3)C=CHCOOH
NCERT Solutions For Class 12 Chemistry Chapter 12 Aldehydes Ketones and Carboxylic Acids-9
Ans.(i) 3 – Phenylpropanoic acid
(ii) 3 – Methylbut-2-enoic acid
(iii)2-Methylcyclohexanecarboxylic acid
(iv)2,4,6 – Trinitrobenzoic acid

12.7 Show how each of the following compounds can be converted into benzoic acid.
(i)Ethylbenzene
(ii)Acetophenone
(iii)Bromobenzene
(iv)Phenylethene (styrene)
Ans.
NCERT Solutions For Class 12 Chemistry Chapter 12 Aldehydes Ketones and Carboxylic Acids-10
NCERT Solutions For Class 12 Chemistry Chapter 12 Aldehydes Ketones and Carboxylic Acids-11

12.8 Which acid of each pair would you expect to be stronger?
(i)CH3CO2H or FCH2CO2H
(ii)FCH2CO2H or ClCH2CO2H
(iii)FCH2CH2CH2CO2H
or CH3CH(F)CH2CO2H
NCERT Solutions For Class 12 Chemistry Chapter 12 Aldehydes Ketones and Carboxylic Acids-12
Ans.
NCERT Solutions For Class 12 Chemistry Chapter 12 Aldehydes Ketones and Carboxylic Acids-13

NCERT EXERCISES

12.1 What is meant by the following terms? Give an example of the reaction in each case.
(i) Cyanohydrin (ii) Acetal (iii) Semicarbazone
(iv) Aldol (v) Hemiacetal (vi) Oxime
(vii) Ketal (viii) Imine (ix) 2,4-DNP derivative
(x) Schiff’s base.
Ans. (i) Cyanohydrin: gem-Hydroxynitriles, i.e., compounds possessing hydroxyl and cyano groups on the same carbon atom are called cyanohydrins. These are produced by addition of HCN to aldehydes or ketones in a weakly basic medium.
NCERT Solutions For Class 12 Chemistry Chapter 12 Aldehydes Ketones and Carboxylic Acids-14
(ii) gem – Dialkoxy compounds in which the two alkoxy groups are present on the terminal carbon atom are called acetals. These are produced by the action of an aldehyde with two equivalents of a monohydric alcohol in presence of dry HCl gas.
NCERT Solutions For Class 12 Chemistry Chapter 12 Aldehydes Ketones and Carboxylic Acids-15
These are easily hydrolysed by dilute mineral acids to regenerate the original aldehydes. Therefore, these are used for the protection of aldehyde group in organic synthesis.
(iii) Semicarbazones are derivatives of aldehydes and ketones and are produced by action of semicarbazide on them in acidic medium.
NCERT Solutions For Class 12 Chemistry Chapter 12 Aldehydes Ketones and Carboxylic Acids-16
(iv) Aldols are P-hydroxy aldehydes or ketones and are produced by the condensation of two molecules of the same or one molecule each of two different aldehydes or ketones in presence of a dilute aqueous base. For example,
NCERT Solutions For Class 12 Chemistry Chapter 12 Aldehydes Ketones and Carboxylic Acids-17
(v)gem – Alkoxyalcohols are called hemiacetals. These are produced by addition of one molecule of a monohydric alcohol to an aldehyde in presence of dry HCl gas.
(vi)Oximes are produced when aldehydes or ketones react with hydroxyl amine in weakly acidic medium.
NCERT Solutions For Class 12 Chemistry Chapter 12 Aldehydes Ketones and Carboxylic Acids-18
(vii) Ketals are produced when a ketone is heated with dihydric alcohols like ethylene glycol in presence of dry HCl gas or /3-toluene sulphonic acid (PTS).
NCERT Solutions For Class 12 Chemistry Chapter 12 Aldehydes Ketones and Carboxylic Acids-19
These are easily hydrolysed by dilute mineral acids to regenerate the original ketones. Therefore, ketals are used for protecting keto groups in organic synthesis.
(viii) Compounds containing -C = N – group are called imines. These are produced when aldehydes and ketones react with ammonia derivatives.
NCERT Solutions For Class 12 Chemistry Chapter 12 Aldehydes Ketones and Carboxylic Acids-20
(ix)2, 4-Dinitrophenyl hydrazone (i.e., 2,4-DNP derivatives) are produced when aldehydes or ketones react with 2,4-dinitrophenyl hydrazine in weakly acidic medium.
NCERT Solutions For Class 12 Chemistry Chapter 12 Aldehydes Ketones and Carboxylic Acids-21
(x) Aldehydes and ketones react with primary aliphatic or aromatic amines to form azomethines or SchifFs bases.
NCERT Solutions For Class 12 Chemistry Chapter 12 Aldehydes Ketones and Carboxylic Acids-22

12.2 Name the following compounds according to IUPAC system of nomenclature:
(i) CH3CH (CH3)—CH2 CH2—CHO (ii) CH3CH2COCH(C2H5)CH2CH2Cl (iii) CH3CH=CHCHO (iv) CH3COCH2COCH3
(v)CH3CH(CH3)CH2C(CH3)2COCH3
(vi)(CH3)3CCH2COOH.
(vii)OHCC6H4CHO-p
Ans. (i)4-Methyl pentanal
(ii)6-Chloro-4-ethylhexan-3-one
(iii)But-2-en-l-al
(iv)Pentane-2,4-dione
(v)3,3,5-Trimethyl-hexan-2-one
(vi)3,3-Dimethyl butanoic acid
(vii)Benzene-1,4-dicarbaldehyde

12.3 Draw the structures of the following compounds:
(i) 3-Methylbutanal
(ii) p-Nitropropiophenone
(iii)p-Methylbenzaldehyde
(iv)4-Methylpent-3-en-2-one
(v)4-Chloropentan-2-one
(vi)3-Bromo-4-phenylpentanoic acid
(vii) pp’-Dihydroxybenzophenone
(viii)Hex-2-en-4-ynoic acid
Ans.
NCERT Solutions For Class 12 Chemistry Chapter 12 Aldehydes Ketones and Carboxylic Acids-23
NCERT Solutions For Class 12 Chemistry Chapter 12 Aldehydes Ketones and Carboxylic Acids-24

12.4 Write the IUPAC names of the following ketones and aldehydes. Wherever possible, give also common names.(i)CH3CO(CH2)4CH3 (ii) CH3CH2CH BrCH2CH(CH3)CHO (iii) CH3(CH2)5CHO (iv) Ph—CH=CH—CHO
NCERT Solutions For Class 12 Chemistry Chapter 12 Aldehydes Ketones and Carboxylic Acids-25
Ans.
NCERT Solutions For Class 12 Chemistry Chapter 12 Aldehydes Ketones and Carboxylic Acids-26

12.5 Draw structures of the following derivatives:
(i)The 2,4-dinitrophenylhydrazone of benzaldehyde
(ii)Cydopropanone oxime
(iii)Acetaldehydedimethylacetal
(iv)The semicarbazone ofcyclobutanone
(v)The ethylene ketal of hexan-3-one
(vi)The methyl hemiacetal of formaldehyde
Ans.
NCERT Solutions For Class 12 Chemistry Chapter 12 Aldehydes Ketones and Carboxylic Acids-27

12.6 Predict the products formed when cyclohexanecarbaldehyde reacts with following reagents. .
(i) PhMgBr and then H3O+
(ii) Tollen reagent
(iii) Semicarbazide and weak acid
(iv)Excess ethanol and acid
(v)Zinc amalgam and dilute hydrochloric acid
Ans.
NCERT Solutions For Class 12 Chemistry Chapter 12 Aldehydes Ketones and Carboxylic Acids-28

12.7Which of the following compounds would undergo aldol condensation, which the Cannizzaro reaction and which neither? Write the structures of the expected products of aldol condensation and Cannizzaro reaction.
(i)Methanal (ii) 2-Methylpentanal (iii) Benzaldehyde .
(iv) Benzophenone (v) Cyclohexanone (vi) 1-Phenylpropanone
(vii) Phenylacetaldehyde (viii) Butan-l-ol 1 (ix) 2,2-Dimethylbutanal
Ans. 2-Methylpertfanal, cyclohexanone, 1-phenylpropanone and phenylacetaldehyde contain one or more a-hydrogen and hence undergo aldol condensation. The reactions and the structures of the expected products are given below:
NCERT Solutions For Class 12 Chemistry Chapter 12 Aldehydes Ketones and Carboxylic Acids-29
NCERT Solutions For Class 12 Chemistry Chapter 12 Aldehydes Ketones and Carboxylic Acids-30

12.8 How will you convert ethanal into the following compounds?
(i)Butane-1,3-diol (ii)But-2-enal (iii)But-2-enoic acid
Ans.
NCERT Solutions For Class 12 Chemistry Chapter 12 Aldehydes Ketones and Carboxylic Acids-31

12.9 Write structural formulas and names of four possible aldol condensation products from propanal and butanal. In each case, indicate which aldehyde acts as nucleophile and which as electrophile.
Ans.
NCERT Solutions For Class 12 Chemistry Chapter 12 Aldehydes Ketones and Carboxylic Acids-32

12.10 An organic compound with the molecular formula C9H10O forms 2,4-DNP derivative, reduces Tollen’s reagent and undergoes Cannizzaro reaction. On vigorous oxidation, it gives 1,2-benzenedicarboxylic acid. Identify the compound. 
Ans. Since the given compound with molecular formula C9H10O forms a 2,4-DNP derivative and reduces Tollen’s reagent, it must be an aldehyde. Since it undergoes Cannizzaro reaction, therefore, CHO group is directly attached to die benzene ring.
Since on vigorous oxidation, it gives 1, 2-benzene dicarboxylic acid, therefore, it must be an ortho- substituted benzaldehyde. The only o-substituted aromatic aldehyde having molecular formula C9H10O is o-ethyl benzaldehyde. Ail the reactions can now be explained on the basis of this structure.
NCERT Solutions For Class 12 Chemistry Chapter 12 Aldehydes Ketones and Carboxylic Acids-33

12.11 An organic compound (A) (molecular formula C8H16O2) was hydrolysed with dilute sulphuric acid to give a carboxylic acid (B} and an alcohol (C). Oxidation of (C) with chromic acid produced (B). (Q on dehydration gives but-l-ene. Write equations for the reactions involved.
Ans.Since an ester A with molecular formula C8H16O2 upon hydrolysis gives carboxylic acid B and the alcohol C and oxidation of C with chromic acid produces the acid B, therefore, both the carboxylic acid B and alcohol C must contain the same number of carbon atoms.
Further, since ester A contains eight carbon atoms, therefore, both the carboxylic acid B and the alcohol C must contain four carbon atoms each.
Since the alcohol C on dehydration gives but-l-ene, therefore, C must be a straight chain alcohol, i.e., butan-l-ol.
If C is butan-l-ol, then the acid B must be butanoic acid and the ester A must be butyl butanoate.The chemical equations are as follows:
NCERT Solutions For Class 12 Chemistry Chapter 12 Aldehydes Ketones and Carboxylic Acids-34

12.12Arrange the following compounds in increasing order of their property as indicated:
(i) Acetaldehyde, Acetone, Di-tert butyl ketone, Methyl tert-butyl ketone (reactivity towards HCN) (ii) CH3CH2CH(Br)COOH, CH3CH(Br)CH2COOH, (CH3)2CH COOH,CH3CH2CH2COOH (acid strength).
(iii) Benzoic acid, 4-Nitrobenzoic acid, 3,4-Dinitrobenzoic add, 4-Methoxybenzok acid (acid strength).
Ans. (i) The reactivity of aldehydes and ketones towards HCN addition decreases as the +1 – effect of the alkyl groups increases. Secondly it decreases with increase in steric hindrance to the nucleophilic attack byCN at the carbonyl carbon. Thus the decreasing order of reactivity towards HCN is,
NCERT Solutions For Class 12 Chemistry Chapter 12 Aldehydes Ketones and Carboxylic Acids-35
(ii)We know that + I-effect decreases while -I-effect increases the acidic strength of carboxylic acids. Since + I-effect of isopropyl group is more than that of propyl group, therefore, (CH3)2CHCOOH is a weaker acid than CH3CH2CH2COOH. Further since -I-effect decreases with distance, therefore CH3CH2CHBrCOOH is a stronger acid than CH3CHBrCH2COOH. Thus, the overall acid strength increases in the order:
NCERT Solutions For Class 12 Chemistry Chapter 12 Aldehydes Ketones and Carboxylic Acids-36
(iii) Since electron-donating groups decreases the acidic strength, therefore, 4-methoxy benzoic acid is a weaker acid than benzoic acid. Further since electron withdrawing groups increase the acidic strength, therefore, both 4-nitrobenzoic acid and 3,4-dinitrobenzoic acid are stronger acids than benzoic acid. Further due to the presence of an additional -NO2 group at /w-position with respect to -COOH group, 3,4-dinitrobenzoic acid is a stronger acid than 4-nitrobenzoic acid. Thus, the overall acidic strength increases in the order:4-methoxy benzoic acid < benzoic acid < 4-nitrobenzoic acid < 3,4-dinitrobenzoic acid.

12.13 Give simple chemical tests to distinguish between the following pairs of compounds.
(i)PropanalandPropanone (ii)Acetophenone and Benzophenone
(iii)Phenol and Benzoic acid (iv)Benzoic acid and Ethyl benzoate
(v)Pentan-2-one and Pentan-3-one (vi)Benzaldehyde and Acetophenone.
(vii)EthanalandPropanal
Ans.
NCERT Solutions For Class 12 Chemistry Chapter 12 Aldehydes Ketones and Carboxylic Acids-37
NCERT Solutions For Class 12 Chemistry Chapter 12 Aldehydes Ketones and Carboxylic Acids-38

12.14 Row will you prepare the following compounds from benzene? You may use any inorganic reagent and any organic reagent having not more than one carbon atom.
(i) Methyl benzoate (ii) m-nitrobenzoic acid (iii) p-nitrobenzoic acid
(iv) Phenylaceticacid (v) p-nitrobenzaldehyde
Ans.
NCERT Solutions For Class 12 Chemistry Chapter 12 Aldehydes Ketones and Carboxylic Acids-39
NCERT Solutions For Class 12 Chemistry Chapter 12 Aldehydes Ketones and Carboxylic Acids-40

12.15 How will you bring about the following conversions in not more than two steps?
(i) PropanonetoPropene
(ii) Benzoic acid to Benzaldehyde
(iii) Ethanol to 3-Hydroxybutanal
(iv) Benzene to m-Nitroacetophenone
(v)Benzaldehyde to Benzophenone –
(vi)Bromobenzeneto 1-PhenylethanoL
(vii) Benzaldehyde to 3-Phenylpropan-1-ol .
(viil) Benzaldehyde to α Hydroxyphenylacetk acid
(ix) Benzoic acid to m-Nitrobenzy 1 alcohol
Ans.
NCERT Solutions For Class 12 Chemistry Chapter 12 Aldehydes Ketones and Carboxylic Acids-41
NCERT Solutions For Class 12 Chemistry Chapter 12 Aldehydes Ketones and Carboxylic Acids-42
NCERT Solutions For Class 12 Chemistry Chapter 12 Aldehydes Ketones and Carboxylic Acids-43

12.16 Describe the following:
(i) Acetylation (ii) Cannizzaro reaction
(iii) Cross aldol condensation (iv) Decarboxylation
Ans. (i) Acetylation refers to the process of introducing an acetyl group into a compound namely, the substitution of an acetyl group for an active hydrogen atom. Acetylation is usually carried out in presence of a base such as pyridine, dimethylanitine, etc.
NCERT Solutions For Class 12 Chemistry Chapter 12 Aldehydes Ketones and Carboxylic Acids-44
(ii)Cannizzaro reaction : Aldehydes which do not contain an a-hydrogen atom, when treated with concentrated alkali solution undergo disproportionation, i.e., self oxidation reduction. As a result, one molecule of the aldehyde is reduced to the corresponding alcohol at the cost of the other which is oxidised to the corresponding carboxylic acid. This reaction is called Cannizzaro reaction.
NCERT Solutions For Class 12 Chemistry Chapter 12 Aldehydes Ketones and Carboxylic Acids-45
(iii) Cross aldol condensation: Aldol condensation between two different aldehydes is called cross aldol condensation.If both aldehydes contain a-hydrogens, It gives a mixture of four products.
NCERT Solutions For Class 12 Chemistry Chapter 12 Aldehydes Ketones and Carboxylic Acids-46
(iv)Decarboxylation: The process of removal of a molecule of CO2 from a carboxylic acid is called decarboxylation. Sodium salts of carboxylic acids when heated with soda-lime undergoes decarboxylation to yield alkanes.
NCERT Solutions For Class 12 Chemistry Chapter 12 Aldehydes Ketones and Carboxylic Acids-47

12.17 Complete each synthesis by giving missing starting material, reagent or products.
NCERT Solutions For Class 12 Chemistry Chapter 12 Aldehydes Ketones and Carboxylic Acids-48
Ans.
NCERT Solutions For Class 12 Chemistry Chapter 12 Aldehydes Ketones and Carboxylic Acids-81
NCERT Solutions For Class 12 Chemistry Chapter 12 Aldehydes Ketones and Carboxylic Acids-82
NCERT Solutions For Class 12 Chemistry Chapter 12 Aldehydes Ketones and Carboxylic Acids-83

12.18 Give plausible explanation for each of the following:
(i) Cyclohexanone forms cyanohydrin in good yield but 2,2, fctrimethylcyclohexanone does not (ii) There are two – NH2 groups in semicarbazide. However, only one is involved in the formation of semicarbazones.
(iii)During the preparation of esters from a carboxylic acid and an alcohol in the presence of an acid catalyst, the water or the ester should be removed as soon as it is formed.
Ans.
NCERT Solutions For Class 12 Chemistry Chapter 12 Aldehydes Ketones and Carboxylic Acids-49
The yield of second reaction is very low because of the presence of three methyl groups at ex-positions with respect to the C = O, the nucleophilic attack by the CN ion does not occur due to steric hinderance. Since there is no such steric hindrance in cyclohexanone, therefore, nucleophilic attack by the CN ion occurs readily and hence cyclohexanone cyanohydrin is obtained in good yield.
NCERT Solutions For Class 12 Chemistry Chapter 12 Aldehydes Ketones and Carboxylic Acids-50
Although semicarbazide has two – NH2 groups but one of them (i.e., which is directly attached to C = O) is involved in resonance as shown above. As a result, electron density on N of this -NH2 group decreases and hence it does not act as a nucleophile. In contrast, the other -NH2 group (i.e.. attached to NH) is not involved in resonance and hence lone pair of electrons present on N atom of this -NH2 group is available for nucleophilic attack on the C = O group of aldehydes and ketones.’
(iii) The formation of esters from a carboxylic acid and an alcohol in presence of an acid catalyst is a reversible reaction.
NCERT Solutions For Class 12 Chemistry Chapter 12 Aldehydes Ketones and Carboxylic Acids-51
Thus to shift the equilibrium in the forward direction, the water or the ester formed should be removed as fast as it is formed.

12.19 An organic compound contains 69-77% carbon, 11-63 % hydrogen and rest oxygen. The molecular mass of the compound is 86. It does not reduce Tottens’ reagent but forms an addition compound with sodium hydrogensulphite and give positive iodoform test. On vigorous oxidation, it gives ethanoic and propanoic acid. Write the possible structure of the compound.
Ans.
NCERT Solutions For Class 12 Chemistry Chapter 12 Aldehydes Ketones and Carboxylic Acids-52
Since the compound form sodium hydrogen sulphite addition product, therefore, it must be either an – aldehyde or methyl/ cyclic ketone. Since the compound does not reduce Tollens’ reagent therefore, it cannot be an aldehyde. Since the compound gives positive iodoform test, therefore, the given compound is a methyl ketone. Since the given compound on vigorous oxidation gives a mixture ofethanoic acid and propanoic acid, therefore, the methyl ketone is pentan-2-one, i.e.,
NCERT Solutions For Class 12 Chemistry Chapter 12 Aldehydes Ketones and Carboxylic Acids-53

12.20 Although phenoxide ion has more number of resonating structures than carboxylate ion, carboxylic acid is a stronger acid than on phenol. Why?
Ans. Consider the resonating structures of carboxylate ion and phenoxide ion.
NCERT Solutions For Class 12 Chemistry Chapter 12 Aldehydes Ketones and Carboxylic Acids-80
In case of phenoxide ion, structures (V – VII) carry a negative charge on the less electronegative carbon atom.Therefore, their contribution towards the resonance stabilization of phenoxide ion is very small.
In structures I and II, (carboxylate ion), the negative charge is delocalized over two oxygen atoms while in structures III and IV, the negative charge on the oxygen atom remains localized only the electrons of the benzene ring are delocalized. Since delocalization of benzene electrons contributes little towards the stability of phenoxide ion therefore, carboxylate ion is much more resonance stabilized than phenoxide ion. Thus, the release of a proton from carboxylic acids is much easier than from phenols. In other words, carboxylic acids are stronger acids than phenols.

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NCERT Solutions For Class 12 Chemistry Chapter 13 Amines

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NCERT Solutions For Class 12 Chemistry Chapter 13 Amines

NCERT INTEXT QUESTIONS

13.1.Classify the following amines as primary, secondary and tertiary:
NCERT Solutions For Class 12 Chemistry Chapter 13 Amines-1
Ans. (i) 1° (ii)-3° (iii) 1° (iv) 2°

13.2.(i)Write the structures of different isomeric amines corresponding to the molecular formula, C4H11N.
(ii)Write 1UPAC names of all the isomers.
(iii)What type of isomerism is exhibited by different pairs of amines?
Ans.
NCERT Solutions For Class 12 Chemistry Chapter 13 Amines-2
Chain isomers: (a) and (c); (a) and (d); (b)and(c); (b)and(d)
Metamers: (e) and (g); (f) and (g) Functional isomers: All 10 amines are functional isomers of 2° and 3° amines and vice-versa.

13.3 How will you convert:
(i)Benzene into aniline
(ii)Benzene into N,N-dimethylaniline
(iii)Cl-(CH2)4-Cl into Hexane -1,6- diamine
Ans.
NCERT Solutions For Class 12 Chemistry Chapter 13 Amines-3

13.4 Arrange the following in increasing order of their basic strength: ‘
(i) C2H5NH2,C6H5NH2,NH3,C6H5CH2NH2 and(C2H5)2 NH
(ii) C2H5NH2,(C2H5)2NH,(C2H5)3N,C6H5NH2
(iii) CH3NH2, (CH3)2NH, (CH3)3N, C6H5NH2, C6H5CH2NH2
Ans.C6H5NH2 < NH3 < C6H5CH2NH2 <  C2H5NH2<(C2H5)2NH
(ii)C6H5NH2 < C2H5NH2 < (C2H5)3N <(C2H5)2NH
(iii) C6H5NH2 < C6H5CH2NH2 < (CH3)3 N < CH3NH2<(CH3)2NH

13.5 Complete the following acid-base reactions and name the products:
(i)CH3CH2CH2NH2+HCl ——–>
(ii)(C2H5)3 N+HCl ——–>
Ans.
NCERT Solutions For Class 12 Chemistry Chapter 13 Amines-4

13.6 Write reactions of the final alkylation product of aniline with excess of methyl iodide in the presence of sodium carbonate solution.
Ans.
NCERT Solutions For Class 12 Chemistry Chapter 13 Amines-5

13.7 Write chemical reaction of aniline with benzoyl chloride and write the name of the product obtained.
Ans.
NCERT Solutions For Class 12 Chemistry Chapter 13 Amines-6

13.8 Write structures of different isomers corresponding to the molecular formula, C3H9N. Write IUPAC names of the isomers which will liberated N2 gas on treatment with nitrons acid.
Ans. In ‘all, four structural isomers are possible. These are:
NCERT Solutions For Class 12 Chemistry Chapter 13 Amines-7

13.9 Convert:
(i)3-Methylanilineinto3-nitrotoluene
(ii)Aniline into 1,3,5- Tribromo benzene
Ans.
NCERT Solutions For Class 12 Chemistry Chapter 13 Amines-8

NCERT EXRECISES

13.1 Write IUPAC names of the following compounds and classify them into primary, secondary and tertiary amines. 
(i) (CH3)2 CHNH2 (ii) CH3(CH2)2NH2 (iii) CH3NHCH(CH3)2
(iv) (CH3)3 CNH2 (v) C6H5NHCH3(vi) (CH3CH2)2NCH3
(vii)m-BrC6H4NH2
Ans. (i) Propan-2-amine(1°)
(ii)Propan-1-amine (1°) ,
(iii)N-Methylpropan-2-amine (2°) .
(iv)2-Methylpropan-2-amine(l°)
(v)N-MethylbenzenamineorN-methylaniline(2°)
(vi)N-Ethyl-N-methylethanamine (3°)
(vii)3-Bromobenzenamine or 3-bromoaniline (1°)

13.2 Give one chemical test to distinguish between the following pairs of compounds:
(i)Methylamine and dimethylamine (ii) Secondary and tertiary amines (iii) Ethylamine and aniline (iv) Aniline and benzylamine (v) Aniline and N-Methylaniline.
Ans.
NCERT Solutions For Class 12 Chemistry Chapter 13 Amines-9
NCERT Solutions For Class 12 Chemistry Chapter 13 Amines-10

13.3 Account for the following
(i)pKb of aniline is more than that of methylamine
(ii) Ethylamine is soluble in water whereas aniline is not.
(iii) Methylamine in water reacts with ferric chloride to precipitate hydrated ferric oxide.
(iv)Although amino group is o and p – directing in aromatic electrophilic substitution reactions, aniline on nitration gives a substantial amount of m-nitroaniline.
(v)Aniline does not undergo Friedel-Crafts reaction.
(vi)Diazonium salts of aromatic amines are more stable than those of aliphatic amines.
(vii)Gabriel phthalimide synthesis is preferred for synthesising primary amines.
Ans. (i) In aniline, the lone pair of electrons on the N-atom is delocalised over the benzene ring.
As a result, electron density on the nitrogen . atom decreases. Whereas in CH3NH2,+ I-effect of -CH3 group increases the electron density on the N-atom. Therefore, aniline is a weaker base than methylamine and hence its pKb value is higher than that of methylamine.
(ii) Ethylamine dissolves in water due to intermolecular H-bonding. However, in case of aniline, due to the large hydrophobic part, i.e., hydrocarbon part, the extent of H-bonding is very less therefore aniline is insoluble in water.
NCERT Solutions For Class 12 Chemistry Chapter 13 Amines-11
(iii)Methylamine being more basic than water, accepts a proton from water liberating OH ions,
NCERT Solutions For Class 12 Chemistry Chapter 13 Amines-12
(iv)Nitration is usually carried out with a mixture of cone HNO3 + cone H2SO4. In presence of these acids, most of aniline gets protonated to form ahilinium ion. Therefore, in presence of acids, the reaction mixture consist of aniline and anilinium ion. Now, -NH2 group in aniline is activating and o, p-directing while the -+NH3 group in anilinium ion is deactivating and rw-directing: Nitration of aniline (due to steric hindrance at o-position) mainly gives p-nitroaniline, the nitration of anilinium ion gives m-nitroaniline. In actual practice, approx a 1:1 mixture of p-nitroaniline and m-nitroaniline is obtained. Thus, nitration of aniline gives a substantial amount of m-nitroaniline due to protonation of the amino group.
NCERT Solutions For Class 12 Chemistry Chapter 13 Amines-13

13.4 Arrange the following:
(i) In decreasing order of pKb values: .
C2H5NH2,C6H5NHCH3,(C2H5)2NH and C6H5NH2
(ii) In increasing order of basic strength:
C6H5NH2, C6H5N(CH3)2, (C2H5)2 NH and CH3NH2.
(iii) In increasing order of basic strength:
(а)Aniline,p-nitroaniline andp-toluidine
(b)C6H5NH2, C6H5NHCH3, C6H5CH2NH2
(iv) In decreasing order of basic strength in gas phase:
C2H5NH2, (C2H5)2NH, (C2H5)3N and NH3
(v) In increasing order of boiling point:
C2H5OH, (CH3)2NH, C2H5NH2
(vi) In increasing order of solubility in water:
C6H5NH2,(C2H5)2NH,C2H5NH2
Ans. (i) Due to delocalisation of lone pair of electrons of the N-atom over the benzene ring,C6H5NH2 and C6H5NHCH3 are far less basic than C2H5NH2 and (C2H,)2NH. Due to +I-effect of the -CH3 group, C6H5NHCH3 is little more basic that C6H5NH2. Among C2H5NH2 and (C2H5)2NH, (C2H5)2NH is more basic than C2H5NHdue to greater+I-effect of two -C2H5 groups. Therefore correct order of decreasing pKb values is:
NCERT Solutions For Class 12 Chemistry Chapter 13 Amines-14
(ii) Among CH3NH2 and (C2H5)2NH, primarily due to the greater +I-effect of the two -C2H5 groups over one -CH3 group, (C2H5)2NH is more basic than CH3NH2.In both C6H5NH2 and C6H5N(CH3)2 lone pair of electrons present on N-atom is delocalized over the benzene ring but C6H5N(CH3)2 is more basic due to +1 effect of two-CH3 groups.
NCERT Solutions For Class 12 Chemistry Chapter 13 Amines-15
(iii) (a) The presence of electron donating -CH3 group increases while the presence of electron withdrawing -NO2 group decreases the basic strength of amines.
NCERT Solutions For Class 12 Chemistry Chapter 13 Amines-16
(b) In C6H5NH2 and C6H5NHCH3, N is directly attached to the benzene ring. As a result, the lone pair of electrons on the N-atom is delocalised over the benzene ring. Therefore, both C6H5NH2 and C6H5NHCH3 are weaker base in comparison to C6H5CH2NH2. Among C6H5NH2 and C6H5NHCH3, due to +1 effect of-CH3 group C6H5NHCH3 is more basic.
NCERT Solutions For Class 12 Chemistry Chapter 13 Amines-17
(iv) In gas phase or in non-aqueous solvents such as chlorobenzene etc, the solvation effects i. e., the stabilization of the conjugate acid due to H-bonding are absent. Therefore, basic strength depends only upon the +I-effect of the alkyl groups. The +I-effect increases with increase in number of alkyl groups.Thus correct order of decreasing basic strength in gas phase is,
NCERT Solutions For Class 12 Chemistry Chapter 13 Amines-18
(v)Since the electronegativity of O is higher than thalof N, therefore, alcohols form stronger H-bonds than amines. Also, the extent of H-bonding depends upon flie number of H-atoms on the N-atom, thus the extent of H-bonding is greater in primary amine than secondary amine.
NCERT Solutions For Class 12 Chemistry Chapter 13 Amines-19
(vi)Solubility decreases with increase in molecular mass of amines due to increase in the size of the hydrophobic hydrocarbon part and with decrease iirthe number of H-atoms on the N-atom which undergo H-bonding.
NCERT Solutions For Class 12 Chemistry Chapter 13 Amines-20

13.5 How wjll you convert:
(i) Ethanoic acid into methanamine (ii) Hexanenitrile into 1-aminopentane
(iii) Methanol to ethanoic acid. (iv) Ethanamine into methanamine
(v) Ethanoic acid into propanoic acid (vi) Methanamine into ethanamine
(vii) Nitromethane into dimethylamine (viii) Propanoic acid into ethanoic acid?
Ans.
NCERT Solutions For Class 12 Chemistry Chapter 13 Amines-21
NCERT Solutions For Class 12 Chemistry Chapter 13 Amines-22

13.6 Describe a method for the identification of primary, secondary and tertiary amines. Also write chemical equations of the reactions involved.
Ans. The three type of amines can be distinguished by Hinsberg test. In this test, the amine is shaken with benzenesulphonyt chloride (C6H5SO2Cl) in the presence of excess of aqueous NaOH or KOH. A primary amine reacts to give a clear solution, which on acidification yields an insoluble compound.
NCERT Solutions For Class 12 Chemistry Chapter 13 Amines-23

13.7 Write short notes on the following:
(i) Carbylamine reaction (il) Diazotisation (iii) ‘Hofmann’s bromamide reaction
(iv) Coupling reaction (v) Ammonolysis (vi) Acetylation
(vii) Gabriel phthalimide synthesis
Ans.(i) Carbylamine reaction: Both aliphatic and aromatic primary amines when warmed with chloroform and an alcoholic solution of KOH, produces isocyanides or carbylamines which have very unpleasant odours. This reaction is called carbylamine reaction.
NCERT Solutions For Class 12 Chemistry Chapter 13 Amines-24
(ii)Diazotisation: The process of conversion of a primary aromatic amino compound into a diazonium salt, is known as diazotisation. This process is carried out by adding an aqueous solution of sodium nitrite to a solution of primary aromatic amine (e.g., aniline) in excess of HCl at a temperature below 5°C.
NCERT Solutions For Class 12 Chemistry Chapter 13 Amines-25
(iii)Hoffmann’s bromamide reaction: When an amide is treated with bromine in alkali solution, it is converted to a primary amine that has one carbon atom less than the starting amide. This reaction is known as Hoffinann’s bromamide degradation reaction.
NCERT Solutions For Class 12 Chemistry Chapter 13 Amines-26
(iv) Coupling reaction: In this reaction, arene diazonium salt reacts with aromatic amino compound (in acidic medium) or a phenol (in alkaline medium) to form brightly coloured azo compounds. The reaction generally takes place at para position to the hydroxy or amino group. If para position is blocked, it occurs at ortho position and if both ortho and para positions are occupied, than no coupling takes place.
NCERT Solutions For Class 12 Chemistry Chapter 13 Amines-27
(v) Ammonolysis: It is a process of replacement of either halogen atom in alkyl halides (or aryl halides) or hydroxyl group in alcohols (or phenols) by amino group. The reagent used for ammonolysis is alcoholic ammonia. Generally, a mixture of primary, secondary and tertiary amine is formed.
NCERT Solutions For Class 12 Chemistry Chapter 13 Amines-28
(vi) Acetylation: The process of introducing an acetyl (CH3CO-) group into molecule using acetyl chloride or acetic anhydride is called acetylation.
NCERT Solutions For Class 12 Chemistry Chapter 13 Amines-29
(vii) Gabriel phthalimide synthesis: It is a method of preparation of pure aliphatic and aralkyl primary amines. Phthalimide on treatment with ethanolic KOH gives potassium phathalimide which on heating with a suitable alkyl Or aralkyl halides gives N-substituted phthalimides, which on hydrolysis with dil HCI or with alkali give primary amines.
NCERT Solutions For Class 12 Chemistry Chapter 13 Amines-30

13.8 Accomplish the following conversions:
(i) Nitrobenzene to benzoic acid (ii) Benzene to m-bromophenol
(iii) Benzoic acid to aniline (iv) Aniline to 2,4,6-tribromofluorobenzene
(v) Benzyl chloride to 2-phenylethanamine (vi) Chlorobenzene to p-Chloroaniline
(vii) Aniline to p-bromoaniIine (viii)Benzamide to toluene
(ix) Aniline to benzyl alcohol.
Ans.
NCERT Solutions For Class 12 Chemistry Chapter 13 Amines-31
NCERT Solutions For Class 12 Chemistry Chapter 13 Amines-32
NCERT Solutions For Class 12 Chemistry Chapter 13 Amines-33

13.9 Give the structures of A,B and C in the following reaction:
NCERT Solutions For Class 12 Chemistry Chapter 13 Amines-34
NCERT Solutions For Class 12 Chemistry Chapter 13 Amines-35
Ans.
NCERT Solutions For Class 12 Chemistry Chapter 13 Amines-36

13.10 An aromatic compound ‘A’on treatment with aqueous ammonia and heating forms compound ‘B’ which on heating with Br2 and KOH forms a compound ‘C’ of molecular formula C6H7N. Write the structures and IUPAC names of compounds A,B and C.
Ans. Since the compound ‘C’ with molecular formula C6H7N is formed from compound ‘B’ on treatment with Br2 KOH, therefore, compound ‘B’ must be an amide and ‘C’ must be an amine.
The only amine having the molecular formula C6H7N, i. e., C6H5NH2 is aniline.
Since ‘C’ is aniline, therefore, die amide from which it is formed must be benzamide (C6H5CONH2). Thus, compound‘B’is benzamide. Since compound ‘B’ is formed from compound ‘A’ with aqueous ammonia and heating, therefore, compound ‘A’ must be benzoic acid.

13.11 Complete the following reactions:
NCERT Solutions For Class 12 Chemistry Chapter 13 Amines-37
Ans.
NCERT Solutions For Class 12 Chemistry Chapter 13 Amines-38
NCERT Solutions For Class 12 Chemistry Chapter 13 Amines-39

13.12 Why cannot aromatic primary amines be prepared by Gabriel phthalimide synthesis?
Ans. The success of Gabriel phthalimide reaction depends upon the nucleophilic attack by the phthalimide anion on the organic halogen compound.
Since aryl halides do not undergo nucleophilic substitution reactions easily, therefore, arylamines, i.e., aromatic, primary amines cannot be prepared by Gabriel phthalimide reaction.
NCERT Solutions For Class 12 Chemistry Chapter 13 Amines-40

13.13 Write the reactions of (i) aromatic and (ii) aliphatic primary amines with nitrous acid.
Ans. Both aromatic and aliphatic primary amines react with HNO2 at 273-278 K to form aromatic and aliphatic diazonium salts respectively. But aliphatic diazonium salts are unstable even at this low temperature and thus decompose readily to form a mixture of compounds. Aromatic and aliphatic primary amines react with
NCERT Solutions For Class 12 Chemistry Chapter 13 Amines-41

13.14 Give plausible explanation for each of the following:
(i) Why are amines less acidic than alcohols of comparable molecular masses?
(ii) Why do primary amines have higher boiling point than tertiary amines?
(iii) Why are aliphatic amines stronger bases than aromatic amines?
Ans. (i) Loss of proton from an amine gives an amide ion while loss of a proton from alcohol give an alkoxide ion.
R—NH2—>R—NH +H+
R—O —H—>R— O +H+ .
Since O is more electronegative than N, so it wijl attract positive species more strongly in comparison to N. Thus, RO~ is more stable than RNH®. Thus, alcohols are more acidic than amines. Conversely, amines are less acidic than alcohols.
(ii) Due to the presence of two H-atoms on N-atom of primary amines, they undergo extensive intermolecular H-bonding while tertiary amines due to the absence of H-atom on the N-atom do not undergo H-bonding. As a result, primary amines have higher boiling points than tertiary amines of comparable molecular mass.
(iii) Aromatic amines are far less basic than ammonia and aliphatic amines because of following reasons:
(a)Due to resonance in aniline and other aromatic amines, the lone pair of electrons on the nitrogen atom gets delocalised over the benzene ring and thus it is less easily available for protonation. Therefore, aromatic amines are weaker bases than ammonia and aliphatic amines.
(b)Aromatic amines arS more stable than corresponding protonated ion; Hence, they hag very less tendency to combine with a proton to form corresponding protonated ion, and thus they are less basic.

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NCERT Solutions For Class 12 Chemistry Chapter 16 Chemistry in Everyday Life

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NCERT Solutions For Class 12 Chemistry Chapter 16 Chemistry in Everyday Life

NCERT INTEXT QUESTIONS

16.1 Sleeping pills are recommended by doctors to the patients suffering from sleeplessness but it is not advisable to take its doses without consultation with the doctor. Why?
Ans. Most of drugs taken in doses higher than recommended may produce harmful effects and act as poison and cause even death. Therefore, a doctor must always be consulted before taking the drug.

16.2 With refrence to which classification has the statement “ranitidine is an antacid”, been given?
Ans. This statement refers to the classification of drugs according to pharmacological effect because any drug which will be used to neutralise the excess acid present in the stomach will be called an antacid.

16.3 Why do we require artificial sweetening agents?
Ans. To reduce calorie intake and to protect teeth from decaying, we need artificial sweeteners.

16.4 Write the chemical equation for preparing sodium soap from glyceryl oleate and glyceryl palmitate. Structures of these compounds are given below:
(i)(C15H31COO)3C3H5-Glyceryl palmitate
(ii)(C17H32COO)3C3H5-Glyceryl oleate
Ans.
NCERT Solutions For Class 12 Chemistry Chapter 16 Chemistry in Everyday Life-1
NCERT Solutions For Class 12 Chemistry Chapter 16 Chemistry in Everyday Life-2

16.5 Following type of non-ionic detergents are present in liquid detergents, emulsifying agents and wetting agents. Label the hydrophilic and hydrophobic part in the molecule. Identify the functional group (s) present in the molecule.
NCERT Solutions For Class 12 Chemistry Chapter 16 Chemistry in Everyday Life-3
Ans.
NCERT Solutions For Class 12 Chemistry Chapter 16 Chemistry in Everyday Life-4
Functional groups present in the detergent molecule are:
(i)ether
(ii)1°alcoholic group

NCERT EXRECISES

16.1 Why do we need to classify drugs in different ways?
Ans.Drugs are classified in following different ways:
(a)Based on pharmacological effect.
(b)Based on action on a particular biochemical process.
(c)Based on chemical structure.
(d)Based on molecular targets.
Each classification has its own usefulness.
(а)Classification based on pharmacological effect is useful for doctors because it provides them the whole range of drugs available for the treatment of a particular disease.
(b)Classification based on action on a particular biochemical proc*ess is useful for choosing the correct compound for designing the synthesis of a desired drug.
(c)Classification based on chemical structure helps us to design the synthesis of a number of structurally similar compounds having different substituents and then choosing the drug having least toxicity.
(d)Classification on the basis of molecular targets is useful for medical chemists so that they can design a drug which is most effective for a particular receptor site.

16.2 Explain the term, target molecules or drug targets as used in medicinal chemistry.
Ans. Drugs interact with macromolecules like proteins, carbohydrates, lipids and nucleic acids thus these macro molecules are called drug targets. These macromolecules perform various functions in the body for example, proteins perform several roles in the body. Proteins which act as biological catalysts are called enzymes, those which are involved in communication system are called receptors. Carrier proteins carry polar molecules across the cell membrane. Nucleic acids have coded genetic information in the cell whereas lipids and carbohydrates form structural part of cell membranes.

16.3 Name the macro molecules that are chosen as drug targets.
Ans. Proteins, carbohydrates, lipids and nucleic acids are chosen as drug targets.

16.4 Why should not medicines be taken without consulting doctors?
Ans. Some drugs can cause side effects when drug binds to more than one type of receptor. Therefore, doctor’s consultation is must to choose the right drug that has the maximum affinity for a particular receptor site to have desired effect. Dose of the drug taken at a time is also crucial because some drugs in higher doses act as poisons and may cause death.

16.5 Define the term chemotherapy.
Ans. It is the branch of chemistry that deals with the treatment of diseases by using chemicals as medicines.

16.6 Which forces are involved in holding the drugs to the active site of enzymes?
Ans.The following forces are involved in holding the drugs to the active site of enzymes:
(a)Hydrogen bonding
(b)Ionic bonding
(c)Dipole-dipole interactions
(d)van der Waals interactions

16.7 While antacids and antiallergic drugs interfere with the function of histamines, why do these not interfere with the function of each other?
Ans. Drugs are designed to cure some ailment in one organ of the body do not affect the other because they work on different receptors. For example, secretion of histamine causes allergy. It also causes acidity due to release of hydrochloric acid in the stomach. Since antiallergic and antacids drugs work on different receptors, therefore, antihistamines remove allergy while antacids remove acidity.

16.8 Low level of noradrenaline is the cause of depression. What type of drugs are needed to cure this problem? Name two drugs.
Ans.In case of low level of neurotransmitter, . noradrenaline, tranquilizer (antidepressant) drugs are required because low levels of noradrenaline leads to depression. These drugs inhibit the enzymes which catalyse the degradation of noradrenaline. If the enzyme is inhibited, noradrenaline is slowly metabolized and can activate its receptor for longer periods of time thereby reducing depression. Two important drugs are iproniazid and phenylzine.

16.9 What is meant by the term broad spectrum antibiotics? Explain.
Ans. Broad spectrum antibiotics are effective against several different types or wide range of harmful bacteria. For example, tetracycline, chloramphenicol and of loxacin. Chloramphenicol can be used in case of typhoid, dysentry, acute fever, urinary infections, meningitis and pneumonia.

16.10 How do antiseptics differ from disinfectants? Give one example of each.
Ans. Antiseptics are chemical substances which prevent the growth of micro-organisms and may even kill them but they are not harmful for human or animal tissues. For example, dettol and savlon. They are generally applied on wounds, cuts, ulcers and diseased skin surfaces. Furacin and soframycin are well known antiseptic creams.
Disinfectants are chemical substances which kill microorganisms but are not safe to be applied to the living tissues. These are generally used to kill microorganisms present in the drains toilets, floors, etc. Some common examples of disinfectants are phenol ( 1% solution) and chlorine (0.2 to 0.4 ppm).

16.11 Why are cimetidine and ranitidine better antacids than sodium hydrogencarbonate or magnesium or aluminium hydroxide?
Ans. If excess of NaHCO3 or Mg(OH)2 or Al(OH)3 is used, it makes the stomach alkaline and thus triggers the release of even more HCl which may cause ulcer in the stomach. In contrast, cimetidine and ranitidine prevent the interaction of histamine with the receptor cells in the stomach wall and thus release of HCl will be less as histamine stimulates the secretion of acid.

16.12 Name a substance which can be used as an antiseptic as well as disinfectant.
Ans. 0.2% solution of phenol acts as antiseptic while 1% solution acts as a disinfectant.

16.13 What are the main constituents of dettol?
Ans. Chloroxylenol .and α-terpineol in a suitable solvent.
NCERT Solutions For Class 12 Chemistry Chapter 16 Chemistry in Everyday Life-5

16.14 What is tincture of iodine? What is its use?
Ans. 2-3% solution of iodine in alcohol and water is called tincture of iodine. It is a powerful antiseptic. It is applied on wounds.

16.15 What are food preservatives?
Ans. Chemical substances which are used to protect food against bacteria, yeasts and moulds are called preservatives. For example, sodium benzoate and sodium metabisulphite.

16.16 Why is the use of aspartame limited to cold foods and drinks?
Ans. This is because it decomposes at baking or cooking temperatures and hence can be used only in cold foods and drinks as an artificial sweetener

16.17 What are artificial sweetening agents? Give two examples.
Ans. Artificial sweeteners are chemical substances which are sweet in taste but do not add any calories to our body. They are excreted as such through urine. For example, saccharin, aspartame, alitame etc.

16.18 Name the sweetening agent used in the preparation of sweets for a diabetic patient.
Ans. Saccharine, aspartame or alitame may be used in the preparation of sweets for a diabetic patient.

16.19 What problem arises in using alitame as artificial sweetener?
Ans. Alitame is a high potency artificial sweetener.Therefore, it is difficult to control the sweetness of the food to which it is added.

16.20 How are synthetic detergents better than soaps?
Ans. They can be used in hard water as well as in acidic solution. The reason being that sulphonic acids and their calcium and magnesium salts are soluble in water thus they do not form curdy white precipitate with hard water but the fatty acids and their calcium and magnesium salts of soaps are insoluble. Detergents also works in slightly acidic solution due to formation of soluble alkyl hydrogen sulphates. Soaps react with acidic solution to form insoluble fatty acids.

16.21 Explain the following terms with suitable examples:
(i) cationic detergents (ii) anionic detergents and (iii) non-ionic detergents
Ans. (i) Cationic detergents: These are quaternary ammonium salts, chlorides, acetates, bromides etc containing one or more long chain alkyl groups. For example, cetyltrimethyl ammonium chloride.
(ii) Anionic detergents are called so because a large part of their molecules are anions. ‘These are of two types:
(a)Sodium alkyl sulphates: For example, sodium lauryl sulphate, C11H23CH2OSO3 Na+.
(b)Sodium alkylbenzenesulphonates.Vor example, sodium 4-(l-dodecyl) benzenesu Iphphonate (SDS).
NCERT Solutions For Class 12 Chemistry Chapter 16 Chemistry in Everyday Life-6
(iii)Neutral or non-ionic detergents: These are esters of high molecular mass alcohols with fatty acids. These can also be obtained by treatment of long chain alcohols by with excess of ethylene oxide in presence of a base. For example, polyethylene glycol stearate,CH3(CH2)16COO (CH2CH2O)11 CH2CH2OH Polyethylene glycol stearate.

16.22 What are biodegradable and non-biodegradable detergents? Give one example of each.
Ans. Detergents having straight chain hydrocarbons are easily degraded (or decomposed) by microorganisms and hence are called biodegradable detergents while detergents containing branched hydrocarbon chains are not easily degraded by the microorganisms find hence are called non-biodegradable detergents. Consequently, non-biodegradable detergents accumulate in rivers and water ways thereby causing severe water pollution. Examples of biodegradable detergents are sodium lauryl sulphate, sodium 4-(-l-dodecyl) benzenesulphonate and sodium 4-(2-dodecyl) benzenesulphonate.
Examples of non-biodegradable detergents is sodium 4-(1, 3,5,7 – tetramethyloctyl) benzene sulphonate.

16.23 Why do soaps not work in hard water?
Ans. Hard water contains calcium and magnesium salts. Therefore, in hard water soaps get precipitated as calcium and magnesium soaps which being insoluble stick to the clothes as gummy mass.

16.24 Can you use soaps and synthetic detergents to check the hardness of water?
Ans. Soaps get precipitated as insoluble calcium and magnesium soaps in hard water but detergents do not. Therefore, soaps but not synthetic detergents can be used to check the hardness of water.

16.25 Explain the cleansing action of soaps.
Ans. Cleansing action of soaps : Soaps contain two parts, a large hydrocarbon which is a hydrophobic (water repelling) and a negative charged head, which is hydrophillic (water attracting). In solution water molecules being polar in nature, surround the ions & not the organic part of the molecule. When a soap is dissolved in water the molecules gather together as clusters, called micelles. The tails stick inwards & the head outwards. The hydrocarbon tail attaches itself to oily dirt. When water is agitated, the oily dirt tends to lift off from the dirty surface & dissociates into fragments. The solution now contains small globules of oil surrounded detergent molecules. The negatively charged heads present in water prevent the small globules from coming together and form aggregates. Thus the oily dirt is removed from the object.

16.26 If water contains dissolved calcium hydrogencarbonate, out of soaps and synthetic detergents, which one will you use for cleaning clothes?
Ans. Calcium hydrogencarbonate makes water hard. Therefore, soap cannot be used because it gets precipitated in hard water. On the other hand, a synthetic detergent does not precipitate in hard water because its calcium salt is also soluble in water. Therefore, synthetic detergents can be used for cleaning clothes in hard water.

16.27 Label the hydrophilic and hydrophobic parts in the following compounds.
(i)CH3(CH2)10CH2OSO3 Na+
(ii)CH3(CH2)15 -N+(CH3)3Br
(iii)CH3(CH2)16C00(CH2CH2O)11CH2CH2OH
Ans.
NCERT Solutions For Class 12 Chemistry Chapter 16 Chemistry in Everyday Life-7

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NCERT Solutions For Class 12 Chemistry Chapter 14 Biomolecules

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NCERT Solutions For Class 12 Chemistry Chapter 14 Biomolecules

NCERT INTEXT QUESTIONS

14.1 Glucose or sucrose are soluble in water but cyclohexane and benzene (simple six membred ring compounds) are insoluble in water Explain.
Ans. The .solubility of a solute in a given solvent follows the rule ‘ Like dissolves like’.Glucose contains five and sucrose contains eight -OH groups. These -OH groups form H-bonds with water. As a result of this extensive intermoleeular H-bonding, glucose and sucrose are soluble in water.On the other hand, benzene and cyclohexane do not contain -OH bonds and hence do not form H-bonds with water. Moreover, they are non-polar molecules and hence do not dissolve in polar water molecules.

14.2 What are the expected products of hydrolysis of lactose?
Ans. Lactose being a disaccharide gives two molecules of monosaccharides Le. one molecule each of D-(+) – glucose and D-(+)-galactbse.
NCERT Solutions For Class 12 Chemistry Chapter 14 Biomolecules-1

14.3 How do you explain the absence of aldehyde group in the pentaacetate of D-glucose?
Ans. The cyclic hemiacetal form of glucose contains an -OH group at C-l which gets hydrolysed in aqueous solution to produce open chain aldehydic form which then reacts with NH2OH -to form corresponding oxime. Thus, glucose contains an aldehydic group. However, when glucose is reacted with acetic anhydride, the -OH group at C-l along with the other -OH groups at C-2, C-3, C-4 and C-6 form a pentaacetate.
Since the penta acetate of1 glucose does not contain a free -OH group at C-l, it cannot get hydrolysed in aqueous solution to produce open chain aldehydic form and hence glucose pentaacetate does not react with NH2OH to form glucose oxime. The reactions are shown as:
NCERT Solutions For Class 12 Chemistry Chapter 14 Biomolecules-2

14.4 The melting points and solubility of amino acids in water are generally higher than those of corresponding haloacids. Explain.
Ans. The amino acids exist as zwitter ions, H3N+ — CHR-COO-. Due to this dipolar salt like character, they have strong dipole-dipole attractions. Therefore, their melting points are higher than corresponding haloacids which do not have salt like character.
Due to salt like character, amino acids intereact strongly with water. As a result, their solubility in water is higher than corresponding haloacids which do not have salt like character.

14.5 Where does the water present in the egg go after boiling the egg?
Ans. When egg is boiled, proteins first undergo denaturation and then coagulation and the water present in the egg gets absorbed in coagulated protein, probably through H- bonding

14.6 Why cannot Vitamin C be stored in our body?
Ans. Vitamin C cannot be stored in the body because it is water soluble and is, therefore, easily excreted in urine.

14.7 What products would be formed when a nucleotide from DNA containing thymine is hydrolysed?
Ans. The products obtained are 2-deoxy-D-ribose, . phosphoric acid and thymine.

14.8 When RNA is hydrolysed, there is no relationship among the quantities of different bases obtained. What does this fact suggest about the structure of RNA?
Ans. A DNA molecule has two strands in which the four complementary bases pair each other, i.e., cytosine (C) always pair with guanine (G) while thymine (T) always pairs with adenine (A). Thus, when a DNA molecule is hydrolysed, the molar amounts of cytosine is always equal to that of guanine and that of adenine is always equal to thymine.In RNA, there is no relationship between the quantities of four bases (C, G, A and U) obtained, therefore, the base pairing principle, i.e. A pairs with U and C pairs with G is not followed. Therefore, unlike DNA, RNA has a single strand.

NCERT EXRECISES

14.1 What are monosaccharides ?
Ans. Monosaccharides are carbohydrates Which cannot be hydrolysed to smaller molecules.Their general formula is (CH2O)n Where n=3-7 These are of two types: Those which contain an aldehyde group (-CHO) are called aldoses and those which contain a keto (C=O) group are called ketoses.
They are further classified as trioses , tetroses ,pentoses , hexoses and heptoses according as they contain 3,4,5,6, and 7 carbon atoms respectively.For example.
NCERT Solutions For Class 12 Chemistry Chapter 14 Biomolecules-3

14.2 What are reducing sugars?
Ans. Carbohydrates which reduces Fehling’s solution to red precipitate of Cu20 or Tollen’s reagent to metallic Ag are called reducing sugars. All monosaccharides (both aldoses and ketoses) and disaccharides except sucrose are reducing sugars. Thus, D – (+) – glucose, D-(-)-fructose, D – (+) – maltose and D – (+) – lactose are reducing sugars.

14.3 Write two main functions of carbohydrates in plants.
Ans. Two major functions of carbohydrates in plants are following
(a)Structural material for plant cell walls: The polysaccharide cellulose acts as the chief structural material of the plant cell walls.
(b)Reserve food material: The polysaccharide starch is the major reserve food material in the plants. It is stored in seeds and act as the reserve food material for the tiny plant till it is capable of making its own food by photosynthesis.

14.4 Classify the following into monosaccharides and disaccharides. Ribose, 2-deoxyribose, maltose, galactose, fructose and lactose.
Ans. Monosaccharides: Ribose, 2-deoxyribose, galactose and fructose. Disaccharides: Maltose and lactose.

14.5 What do you understand by the term glycosidic linkage?
Ans. The ethereal or oxide linkage through which two monosaccharide units are joined together by the loss of a water molecule to form a molecule of disaccharide is called the glycosidic linkage. The glycosidic linkage in maltpse molecule is shown below:
NCERT Solutions For Class 12 Chemistry Chapter 14 Biomolecules-4

14.6 What is glycogen? How is it different from starch?
Ans. Glycogen is a condensation polymer of α-D glucose. Starch is not a single compound but is a mixture of two components—a water soluble component called amyldse (15- 20%) and water insoluble component amylopectin (80 – 85%). Amylose is a linear polymer of α – D – glucose. But both glycogen and amylopectin are branched polymers of α – D – glucose; father glycogen is more highly branched than amylopectin as amylopectin chains consists of 20 – 25 glucose units, glycogen chains consist of 10 – 14 glucose units.

14.7 What are the hydrolysis products of (i) sucrose, and (ii) lactose?
Ans. Both sucrose and lactose are disaccharides. Sucrose on hydrolysis gives one molecule each of glucose and fructose but lactose on hydrolysis gives one molecule each of glucose and galactose.
NCERT Solutions For Class 12 Chemistry Chapter 14 Biomolecules-5

14.8 What is the basic structural difference between starch and cellulose?
Ans. Starch consists of amylose and amylopectin. Amylose is a linear polymer of α-D-glucose while cellulose is a linear polymer of β -D- glucose. In amylose, C -1 of one glucose unit is connected to C – 4 of the other through α-glycosidic linkage. However in cellulose, C – 1 of one glucose unit is connected to C-4 of the other through β – glycosidic linkage. Amylopectin on the other hand has highly branched structure.
NCERT Solutions For Class 12 Chemistry Chapter 14 Biomolecules-6

14.9 What happens when D-glucose is treated with . the following reagents.
(i)HI (ii) Bromine water (iii) HNO3
Ans.
NCERT Solutions For Class 12 Chemistry Chapter 14 Biomolecules-7

14.10 Enumerate the reactions of D-glucose which cannot be explained by its open chain structure.
Ans. (a) D (+) – glucose does not undergo certain characteristic reactions of aldehydes, e.g., glucose does not form NaHSO3 addition product.
(b)Glucose reacts with NH2OH to form an oxime but glucose pentaacetate does not. This implies that the aldehydic group is absent in glucose pentaacetate.
(c)D – (+) – glucose exists in two stereoisomeric forms, i.e., α -glucose and β-glucose.
(d)Both α – D – glucose and β – D – glucose undergo mutarotation in aqueous solution. Although the crystalline forms of α-  and β -D (+) – glucose are quite stable in aqueous solution but each form slowly changes into an equilibrium mixture of both.
(e)D (+) – glucose forms two isomeric methyl glucosides. Aldehydes normally react with two moles of methanol per mole of the aldehyde to form an acetal but D (+) – glucose when treated with methanol in presence of dry HCl gas, reacts with only one mole of methanol per mole of glucose to form a mixture of two methyl D – glucosides i. e., methyl – α – D – glucoside (melting point 43 8 K, specific rotation +158°) and methyl – β – D – glucoside (melting point 308 K, specific rotation – 33°).

14.11 What are essential and non-essential amino acids? Give two examples of each type.
Ans. α-Amino acids which are needed for good health and proper growth of human beings but are not synthesized by the human body are called- essential amino acids. For example, valine, leucine, phenylalanine, etc. On the other hand, α-amino acids which are needed for health and growth of human beings and are synthesized by the human body are called non-essential amino acids. For example, glycine, alanine, aspartic acid etc.

14.12 Define the following as related to proteins:
(i)Peptide linkage
(ii)Primary structure
(iii)Denaturation
Ans. (i) Peptide bond: Proteins are condensation polymers of α-amino acids in which the same or different α-amino acids are joined by peptide bonds. Chemically, a peptide bond is an amide linkage formed between – COOH group of one α-amino acid and -NH-, group of the other α-amino acid by loss of a molecule of water. For example,
NCERT Solutions For Class 12 Chemistry Chapter 14 Biomolecules-8
(ii) Primary structure: Proteins may contain one or more polypeptide chains. Each . polypeptide chain has a large number of α-amino acids which are linked to one another in a specific manner. The specific sequence in which the various amino acids present in a protein linked to one another is called its primary structure. Any change in the sequence of α-amino acids creates a different protein.
NCERT Solutions For Class 12 Chemistry Chapter 14 Biomolecules-9
(iii) Denaturation: Each protein in the biological system has a unique three-dimensional structure and has specific biologicalactivity. This is called native form of a protein. When a protein in its native form is subjected to a physical change such as change in temperature or a chemical change like change in pH, etc., hydrogen bonds gets broken. As a result, soluble forms of proteins such as globular proteins undergo coagulation or precipitation to give fibrous proteins which are insoluble in water. This coagulation also results in loss of biological activity of the proteins and this loss in biological activity, is called denaturation. During denaturation, 2° and 3° structures of proteins are destroyed but 1° structure remains intact.
The most common example of denaturation of proteins is the coagulation of albumin present in the white of an egg. When the egg is boiled hard, the soluble globular protein present in it is denatured and is converted into insoluble fibrous protein.

14.13 What are the common types of secondary structure of proteins?
Ans. The conformation which the polypeptide chains assume as a result of hydrogen bonding is called secondary structure of the proteins. The two types of secondary structures are α-helix and β-pleated sheet structure.

14.14 What type of bonding helps in stabilising the α-helix structure of proteins?
Ans. The α-helix structure of proteins is stabilized by intramolecular H-bonding between C = O of one amino acid residue and the N – H of the fourth amino acid residue in the chain. This causes the polypeptide chain to coil up into a spiral structure called right handed α- helix structure.

14.15 Differentiate between globular and fibrous proteins.
Ans. (i) Fibrous proteins: These proteins consist of linear thread like molecules which tend to lie side by side (parallel) to form fibres. The polypeptide chains in them are held together usually at many points by hydrogen bonds and some disulphide bonds. As a result,intermolecular forces of attraction are very’ strong and hence fibrous proteins are insoluble in water. Further, these proteins are stable to moderate changes in temperature and pH. Fibrous proteins serve as the chief structural material of animal tissues.For example, keratin in skin, hair, nails and wool, collagen in tendons, fibrosis in silk and myosin in muscles.
(ii) Globular proteins: The polypeptide chain in these proteins is folded around itself in such a way so as to give the entire protein molecule an almost spheroidal shape. The folding takes place in such a manner that hydrophobic (non-polar) parts are pushed inwards and hydrophilic (polar) parts are pushed outwards. As a result, water molecules interact strongly with the polar groups and hence globular protein are water soluble. As compared to fibrous proteins, these are very sensitive to small changes of temperature and pH. This class of proteins include all enzymes, many hormones such as insulin from pancreas, thyroglobulin from thyroid gland, etc.

14.16 How do you explain the amphoteric behaviour of amino acids?
Ans. Amino acids contain an acidic (carboxyl group) and basic (amino group) group in the same molecule. In aqueous solution, they neutralize each other. The carboxyl group loses a proton while the amino group accepts it. As a result, a dipolar or zwitter ion is formed.
NCERT Solutions For Class 12 Chemistry Chapter 14 Biomolecules-10
In zwitter ionjc form, a-amino acid show amphoteric behaviour as they react with both acids and bases.
NCERT Solutions For Class 12 Chemistry Chapter 14 Biomolecules-11

14.17 What are enzymes?
Ans. Enzymes are biological catalyst. Each biological reaction requires a different enzyme. Thus, as compared to conventional catalyst enzymes are very specific and efficient in their action. Each type of enzyme has its own specific optimum conditions of concentration, pH and temperature at which it works best.

14.18 What is the effect of denaturation on the structure of proteins?
Ans. During denaturation, 2° and 3° structures of proteins are destroyed but 1° structure remains intact. As a result of denaturation, die globular proteins (soluble in H2O) are converted into fibrous proteins (insoluble in H2O) and their biological activity is lost. For example, boiled egg which contains coagulated proteins cannot be hatched.

14.19 How are vitamins classified? Name the vitamin responsible for the coagulation of blood.
Ans. Vitamins are classified into two groups depending upon their solubility in water or fat: (i) Water soluble vitamins: These include vitamin B-complex (B1, B2, B5, i.e., nicotinic acid,B6, B12, pantothenic acid, biotin, i.e., vitamin H and folic acid) and vitamin C.
(ii) Fat soluble vitamins: These include vitamins A, D, E and K. They are stored in liver and adipose (fat storing) tissues. Vitamin K is responsible for coagulation of blood.

14.20 Why are vitamin A and vitamin C essential to us? Give their important sources.
Ans. Vitamin A is essential for us because its deficiency causes xerophthalmia (hardening of cornea of eye) and night blindness.
Sources: Fish liver oil, carrots, butter, milk, etc. Vitamin C is essential for us because its deficiency causes scurvy (bleeding of gums) and pyorrhea (loosening and bleeding of teeth). Sources: Citrous fruits, amla, green leafy vegetables etc.

14.21 What are nucleic acids ? Mention their two important functions.
Ans. Nucleic acids are biomolecules which are found in the nuclei of all living cell in form of nucleoproteins or chromosomes (proteins contains nucleic acids as the prosthetic group).
NCERT Solutions For Class 12 Chemistry Chapter 14 Biomolecules-12
Nucleic acids are of two types: deoxyribonucleic acid (DNA) and ribonucleic acid.(RNA).
The two main functions of nucleic acids are:
(a) DNA is responsible for transmission of hereditary effects from one generation to another. This is due to its unique property of replication, during cell division and two identical DNA strands are transferred to the daughter cells.
(b) DNA and RNA are responsible for synthesis of all proteins needed for the growth and maintenance of our body. Actually the proteins are synthesized by various RNA molecules (r-RNA, m-RNA) and t-RNA) in the cell but the message for the synthesis of a particular protein is coded in DNA.

14.22 What is the difference between a nucleoside and a nucleotide?
Ans. A nucleoside contains only two basic components of nucleic acids i.e., a pentose sugar and a nitrogenous base. It is formed when 1- position of pyrimidine (cytosine, thiamine or uracil) or 9-position of purine (guanine or adenine) base is attached to C -1 of sugar (ribose or deoxyribose) by a β-linkage. Nucleic acids are also called polynucleotides since the repeating structural unit of nucleic acids is a nucleotide.
A nucleotide contains all the three basic . components of nucleic acids, i.e., a phosphoric acid group, a pentose sugar and a nitrogenous base. These are obtained by esterification of C5, – OH group of the pentose sugar by phosphoric acid.
NCERT Solutions For Class 12 Chemistry Chapter 14 Biomolecules-13

14.23 The two strands in DNA are not identical but are complementary. Explain.
Ans. The two strands in DNA molecule are held together by hydrogen bonds between purine base of one strand and pyrimidine base of the other and vice versa. Because of different sizes and geometries of the bases, the only possible pairing in DNA are G (guanine) and C (cytosine) through three H-bonds, (i.e.,C = G) and between A (adenine) and T (thiamine) through two H-bonds (i.e., A = T). Due to this base -pairing principle, the sequence of bases in one strand automatically fixes the sequence of bases in the other strand. Thus, the two strands are complimentary and not identical.

14.24 Write the important structural and functional differences between DNA and RNA.
Ans.
NCERT Solutions For Class 12 Chemistry Chapter 14 Biomolecules-14

14.25 What are the different types of RNA found in the cell?
Ans. There are three types of RNA:
(a) Ribosomal RNA (r RNA) (b) Messenger RNA (m RNA) (c) Transfer RNA (t RNA)

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NCERT Exemplar Class 11 Biology Solutions Transport in Plants

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NCERT Exemplar Class 11 Biology Solutions Transport in Plants

Multiple Choice Questions
1.Which of the following statements does not apply to reverse osmosis?
(a) it is used for water purification.
(b) In this technique, pressure greater than osmotic pressure is applied to the system.
(c) It is a passive process.
(d) It is an active process.
Soln. (c) : Reverse osmosis is also known as hyperfiltration. It is a technique that allows the removal of water molecules from various contaminants through a semi-permeable membrane. The process requires a driving force, viz, pressure from a pump, to push the fluid through the membrane. Reverse osmosis is used in removing salts from saline water and extra purification of water.

2.Which one of the following will not directly affect transpiration?
(a) Temperature
(b) Light
(c) Wind speed
(d) Chlorophyll content of leaves
Soln. (d): External factors affecting the rate of transpiration are : atmospheric temperature, relative humidity, light, wind velocity, atmospheric pressure and available soil water. Some internal factors also affect rate of transpiration e.g., leaf area, leaf structure and age of plants. Chlorophyll content of leaves does not directly affect rate of transpiration.

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3.The lower surface of leaf will have more number of stomata in a
(a) dorsiventral leaf
(b) isobilateral leaf
(c) both (a) and (b)
(d) none of these.
Soln.(a): Transpiration is the evaporative loss of water by plants. It occurs mainly through the stomata in the leaves. Usually the lower surface of a dorsiventral (often dicotyledonous) leaf has a greater number of stomata while in an isobilateral (often monocotyledonous) leaf, there c,re almost equal number of stomata on both surfaces.

4.The form of sugar transported through phloem is
(a) glucose (b) fructose
(c) sucrose (d) ribose.
Soln.(c): The sugars, synthesised in leaves (as a result of photosynthesis) are translocated downwards, upwards and laterally to all the other organs including storage organs mainly through phloem. These sugars are translocated in the form of sucrose.

5.The process of guttation takes place
(a) when the root pressure is high and the rate of transpiration is low
(b) when the root pressure is low and the rate of transpiration is high
(c) when the root pressure equals the rate of transpiration
(d) when the root pressure as well as rate of transpiration are high.
Soln.(a): The process of exudation of liquid drops from the edges of leaves is called guttation. Usually it occurs though special structures, called hydathodes. It is the process by which fully turgid plants remove extra water. The cause of guttation is mainly root pressure (when the root pressure is high and rate of transpiration is low, guttation takes place).

6.Which of the following is an example of imbibition?
(a) Uptake of water by root hair
(b) Exchange of gases in stomata
(c) Swelling of seed when put in soil
(d) Opening of stomata
Soln.(c) : Imbibition is a process in which water is absorbed by solid particles causing them to enormously increase in volume without forming a solution. The classical examples of imbibition are absorption of water by seeds and dry wood.

7.When a plant undergoes senescence, the nutrients may be
(a) accumulated
(b) bound to cell wall
(c) translocated
(d) none of these.
Soln.(d) : In case of perennial plants, plant parts undergo senescence and nutrient resources (minerals and carbohydrates) shift away or are translocated from the senescened part to other parts. In leaves at the time of senescence we observe loss of chlorophyll, RNA and protein, transport of nutrients and then abscission, but in case of annual plants, whole plant undergoes senescence and nutrients are used up and no translocation, accumulation and bounding to cell wall takes place. ’

8.Water potential of pure water at standard temperature is equal to
(a) 10 (b) 20
(c) Zero(d) none of these.
Soln.(c)

9.Choose the correct option. Mycorrhiza is a symbiotic association of fungus with root system which helps in
A.Absorption of water
B.Mineral nutrition
C.Translocation
D.Gaseous exchange
(a) Only A
(b) On!yB
(c) both A and B
(d) both B and C
Soln.(c) : The roots of some higher plants are associated with fungal mycelia. Such an association is called mycorrhiza. It represents a mutualistic (symbiotic) association between the root system of higher plants and fungal hyphae. The fungal hyphae form a network around the young root and invade the root cells. The external hyphae greatly increase the surface area for absorption of minerals and water from a much larger volume of soil that perhaps the roots alone cannot do.

10.Based on the figure given below which of the following statements is not correct?
ncert-exemplar-class-11-biology-solutions-transport-in-plants-1
(a) Movement of solvent molecules will take place from chamber A to B
(b) Movement of solute will take place from A to B
(c) Presence of a semipermeable is a pre-requisite for this process to occur.
(d) The direction and rate of osmosis depends on both the pressure gradient and concentration gradient.
Soln.(b) : Solution A and solution B are separated by a semi-permeable membrane. Solution A has higher water potential. Therefore, osmotic movement of water molecules (solvent molecules) will occur from solution A to B. Solute particles cannot pass through semi-permeable membrane.

11.Match the following and choose the correct option.
A.Leaves                     (i) Anti-transpirant
B.Seed                         (ii)Transpiration
C.Roots                      (iii)Negative osmotic potential
D.Aspirin                  (iv) Imbibition
E.Plasmolysed cell (v) Absorption Options:
(a) A-(ii), B-(iv), C-(v), D-(i), E-(iii)
(b) A-(iii), B-(ii), C-(iv), D-(i), E-(v)
(c) A-(i), B-(ii), C-(iii),D-(iv), E-(v)
(d) A-(v), B-(iv), C-(iii),D-(ii), E-(i)
Soln.(a)

12. Mark the mismatched pair.
(a) Amyloplast – Store protein granule
(b) Elaioplast – Store oils or fats
(c) Chloroplasts- Contain chlorophyll pigments
(d) Chromoplasts – Contain coloured pigments other than chlorophyll
Soln.(a) : Amyloplasts are non pigmented organelles found in some plant cells. They are responsible for synthesis and storage of starch granules, through polymerisation of glucose.

Short Answer Type Questions
1.Smaller, lipid soluble molecules diffuse faster
through cell membrane, but the movement of hydrophilicsubstancesare facilitated by certain transporters which are chemically________
Soln. Smaller, lipid soluble molecules diffuse faster through cell membrane, but the movement of hydrophilic substances are facilitated by certain transporters which are chemically proteins.

2.In a passive transport across a membrane,
when two protein molecules move in opposite direction and independent of each other, it is called as ________
Soln.ln a passive transport across a membrane, when two protein molecules move in opposite direction and independent of each other, it is called as diffusion.

3.Osmosis is a special kind of diffusion, in which
water diffuses across the cell membrane. The rate and direction of osmosis depends upon both________
Soln. Osmosis is a special kind of diffusion, in which water diffuses across the cell membrane.
The rate and direction of osmosis depends upon both pressure and concentration gradient.

4.A flowering plant is planted in an earthen pot and irrigated. Urea is added to make the plant grow faster, but after sometime the plant dies. This may be due to___________
Soln. A flowering plant is planted in an earthen pot and irrigated. Urea is added to make the plant grow faster, but after some time the plant dies. This may be due to exosmosis.

5.Absorption of water from soil by dry seeds increases the___________ thus helping seedlings to come out of soil
Soln. Absorption of water from soil by dry seeds increases the pressure, thus helping seedlings to come out of soil. During germination of seedling imbibition pressure develops. .

6.Water moves up against gravity and even for a tree of 20 m height, the tip receives water within two hours. The most important physiological phenomenon which is responsible for the upward movement of water is___________ .
Soln.Water moves up against gravity and even for a tree of 20 m height, the tip receives water within two hours. The most important physiological phenomenon which is responsible for the upward movement of water is transpiration pull.

7.The plant cell cytoplasm is surrounded by both cell wall and cell membrane. The specificity of transport of substances are mostly across the cell membrane, because___________
Soln. The plant cell cytoplasm is surrounded by both cell wall and cell membrane. The specificity of transport of substances are mostly across the cell membrane, because cell wall is freely permeable while cell membrane is selectively permeable, for solutes.

8.TheC4 plants are twice as efficient as C3 plants in terms of fixing C02 but lose only___________ as much water as C3 plants for the same amount ofC02 fixed.
Soln. The C4 plants are twice as efficient as C3 plants in terms of fixingC02 but lose only half as much water asC3plants for the same amount of C02 fixed.

9.Movement of substances in xylem is unidirec-tional while in phloem it is bidirectional. Explain.
Soln. Direction of transport is determined by the sink or area of utilisation of the material under transport.
Xylem transport is unidirectional and up-wards from roots to shoot tips. Phloem trans-port food from source to sink where source is the part of plant which stores or synthesises food and sink is the part which needs or consumes food. These source and sink parts of a plant vary depending on the season or need of the plants, thus the food needs to travel in both upward and downward direction. So, phloem shows bidirectional movement of substance.

10.Identify the processes occurring in I, II and III
ncert-exemplar-class-11-biology-solutions-transport-in-plants-2
Soln.
I – Uniport
II – Antiport
III – Symport

11 .Given below is a table. Fill in the gaps.
.ncert-exemplar-class-11-biology-solutions-transport-in-plants-10
ncert-exemplar-class-11-biology-solutions-transport-in-plants-11
Soln.
ncert-exemplar-class-11-biology-solutions-transport-in-plants-3

12.Define water potential and solute potential.
Soln. Water potential (ψw ) is difference in chemical potential per molar volume of water in a system and that of pure water at the same temperature and pressure.Solute potential (ψs) is the decrease in chemical. potential of pure water due to presence of solute particles in it.

13. Why is solute potential always negative? Explain ψw = ψs + ψp
Soln. Water potential is zero for pure water. Addition of solutes further decreases the potential and therefore, chemical potential is always negative or less than zero,
ψw = ψs + ψp   
ψw= Water potential of a system
ψs = Solute potential (Value negative)
ψ = Pressure potential (Value positive)

14. An onion peel was taken and
(a) Placed in salt solution for five minutes.
(b) After that it was placed in distilled water. When seen under the microscope what would be observed in (a) and (b)?
Soln.In condition (a) plasmolysis or shrinkage of protoplasm away from the cell wall will be observed.
In condition (b) deplasmolysis or swelling of shrunken protoplasm will be observed. As a result, the protoplasm will again come in contact with cell wall

15.Differentiate between apoplast and symplast pathways of water movement. Which of these would need active transport?
Soln. In apoplast pathway, non-living parts of plant body are involved as passage way i.e., cell walls and intercellular spaces.
In symplast pathway living parts of plant body are involved as passage way i.e., protoplasts connected by plasmodesmata.
Symplast pathways of water movement need active transport.

16.How does most of the water moves within the root?
Soln. Most of the water moves passively in the root through the apoplast except in the region of functional endodermis because the casparian strip present in endodermis blocks the apoplastic pathway and water has to enter the symplast.

17. Give the location of Casparian strip and explain its role in the water movement.
Soln. Casparian strips are found in the endodermal cell walls of plant roots. It prevents water movement from returning to the cortex from pericycle, consequently a positive hydrostatic pressure is established in vascular tissue.

18. Differentiate between guttation and transpiration.
Soln. Differences between guttation and transpiration are as follows.
ncert-exemplar-class-11-biology-solutions-transport-in-plants-4

19. Transpiration is a necessary evil in plants. Explain.
Soln. Transpiration causes loss of huge amount of water absorbed by plants and leads to wilting and injury in plants. It also checks photosynthesis, reduces growth and if it is too severe, it may cause death by desiccation. Inspite of various detrimental disadvantages the plants cannot avoid transpiration due to their peculiar structure of leaves which is basically meant for gaseous exchange during respiration and photosynthesis. Besides this, transpiration provides the necessary pull required for ascent of sap. Therefore, transpiration is also regarded as “necessary evil” by Curtis (1926) or “unavoidable evil” by Steward (1959).

20. Describe briefly the three physical properties of water which help in ascent of water in xylem.
Soln.Three properties of water essential for its ascent in xylem are cohesive or cohesion force, adhesive or adhesion force and surface tension.
(i) Cohesion force: It is the force by which water molecules remain attached to one another. The force develops due to hydrogen bonds amongst the water molecules. Cohesion provides tensile strength against negative pressure exerted by transpiration pull.
(ii) Adhesion force: It is the force of attraction between water molecules and inner surface of xylem channels. Adhesion force adds tc tensile strength of water.
(iii) Surface tension: It refers to the higher attraction present amongst water molecules at the interphase between liquid and gaseous phase.

21.A gardener forgot to water a potted plant for a day during summer, what will happen to the plant? Do you think it is reversible? If yes, how?
Soln.Yes, watering of potted plant in summer for a day will cause partial water deficit in the plant. It will result in loss of turgidity in softer tissue. The plant will show wilting or shrivelled and droop appearance of leaves. The condition is reversible if it is not watered only for a day. As the plant is watered next
day, the softer tissue will again become turgid and the leaves will appear stretched again.

22.Identify a type of molecular movement which is highly selective and requires special membrane proteins, but does not require energy.
Soln. Facilitated diffusion is a type of movement which is highly selective and requires special membrane protein, but does not require energy.

23. Correct the statements.
(a) Cells shrink in hypotonic solutions and swell in hypertonic solutions.
(b) Imbibition is a special type of diffusion when water is absorbed by living cells.
(c) Most of the water flow in the roots occurs via the symplast.
Soln.(a) Cells shrink in hypertonic solution and swell in hypotonic solution.
(a) Imbibition is a special type of diffusion when water is adsorbed by non-living parts.
(b) Most of the water flow in the roots occur via the apoplast.

Short Answer Type Questions
1.Minerals absorbed by the roots travel up the xylem. How do they reach the parts where they are needed most? Do all the parts of the plant get the same amount of the minerals?
Soln. Minerals absorbed are transported up the stem, to all parts of plant through the transpirational stream. They are needed the most in the growing region of the plant such as the apical meristem, young leaves, developing flowers, fruits, seeds and the storage organs. All parts do not require same amount of the minerals therefore they do not get the same amount of minerals. Minerals exit the xylem and enter the cells where they are needed by the mechanisms similar to ones by which they entered the xylem. Concentration of minerals is lower in the sink cells than that in the xylem sap, thus these minerals enter them passively. In some cells minerals already present in higher concentration may be taken in by active transport.

2.If one wants to find minerals and in the form they are mobilised in the plant, how will an analysis of the exudate help?
Soln.Exudate (plant sap) consists of xylem sap and small quantity of phloem sap. It is mixture of the organic/inorganic compounds and ions, minerals, sugar and amino acids.
Chemical analysis of plant sap will indicate which, mineral nutrient is transported in –
which form in the plants, e.g., nitrogen is absorbed and transported asN02  and N0 3 and sulphur as sulphate ion forms.

3.From your knowledge of physiology can you think of some method of increasing the life of cut plants in a vase?
Soln.Method of increasing the life of cut plants in a vase are as follows:
(i) Immerse the bases of cut plants immediate ly  in water to prevent entry of air in the xylem channels i.e. cavitation. :
(ii)Adding a small quantity of cytokinin in water of the vase. Cytokinin delays the senescence of plant.

4.Do different species of plants growing in the same area show the same rate of transpiration at a particular time? Justify your answer.
Soln.Different species of plants growing
in the same area do not show the same rate of transpiration at a particular time. This is
because rate of transpiration depends upon various factors like the leaf area, thickness of cuticle, number and duration of opening of stomata and the amount of water absorption etc., and these factors are specific for each type of plant.

5.Water is indispensable for life. What properties of water make it useful for all biological processes on the earth?
Soln. Properties of water which make it useful for all biological processes on the earth are as follows:
(i) Water is a polar molecule which has both positive (hydrogen) and negative (oxygen) ends. Polarity of water molecules makes
water an excellent solvent. That is why, water forms 90% of the protoplasm and provides a medium for biochemical reactions.
(ii)Due to its solvent properties and strong adhesive and cohesive forces water serves as the means to transport, nutrients upwards (ascent of sap) to the shoot system in plants
(iii)Compared to other liquids water requires a relatively large energy input to raise its temperature. This large energy input requirement is important for organisms because it helps to buffer temperature fluctuations
(iv)The high latent heat of vaporisation of water enables organisms to cool themselves by evaporation of water (i.e. transpiration and sweating).

6.How is it that the intracellular levels of K are higher than extracellular levels in animal cells?
Soln. Animal cells operate ATP-energised   Na+-K pump. It sends out 3Na+ ions into extracellular fluid and 2 K ions from extracellular fluid pass into inside of cells. As a result, intracellularK levels become higher than extracellular levels.

7.Cut pieces of beetroot do not leave colour in cold water but do so in hot water. Explain.
Soln. Cell membranes rupture at high temperatures. As a result the pigment contained in cell vacuoles of beetroot pieces leak out.

8.In a girdled plant, when water is supplied to the leaves above the girdle, leaves may remain green for some time then wilt and ultimately die. What does it indicate?
Soln.In a girdled plant, when water is supplied to the leaves above the girdle, leaves may remain green for some time because leaves can synthesise their food by photosynthesis. But, roots not only supply water to the leaves, they supply minerals too. Even when water is supplied to the leaves, they will not get minerals. Minerals are required for photolysis of water during photosynthesis (Mn), for completion of various reactions as they act as cofactors for various enzymes, for integrity of certain ribosomes (Mg) and production of chlorophyll (Mg, N) etc. Thus, leaves of girdled plant will die even when supplied with water due to deficiency of minerals.

9.Various types of transport mechanisms are needed to fulfil the mineral requirements of a plant. Why are they not fulfilled by diffusion alone?
Soln. Diffusion is very slow process for movement of non gaseous substances and allows movement of molecules for short distance only that too along the concentration gradient. In higher plants, requirements of minerals is quite high in plants due to contihued photosynthesis and growth which is not fulfilled by diffusion alone besides, some minerals have to be transported against the concentration gradient. Therefore plants use various channels for mineral transport like facilitated diffusion and active transport.

10.How can plants be grown under limited water supply without compromising on metabolic activities?
Soln. Plants can be grown under limited water supply without compromising on metabolic activities by using antitranspirants e.g. besides, some minerals have to be transported against PMA (Phenyl Mercuric Acetate) ABA (Abscisic Acid). Antitranspirants reduce water loss without affecting diffusion of gases and metabolic activities of the plants.

11.Will the ascent of sap be possible without the cohesion and adhesion of the water molecules? Explain.
Soln.No, the ascent of sap is not possible without cohesion and adhesion of the water molecules. In ascent of sap, continuous column of water is required which is maintained by cohesive and adhesive properties of water molecules. These together provide a very high tensile strength to the water column in xylem which prevents cavitation of the xylem vessels and the breaking of water column that may occur due to continuous transpiration.

12.Keep some freshly cut flowers in a solution of food colour. Wait for some time for the dye to rise in the flower, when the stem of the flower is held up in light, coloured strands can be seen inside. Can this experiment demonstrate which tissue is conducting water up the stem?
Soln. Coloured strands indicate that the food colour has passed upward through a conducting channel. This conduction channel can be xylem. This can be confirmed through microscopic observation of T. S. of stem.

13.When a freshly collected Spirogyra filament is kept in a 10% potassium nitrate solution, it is observed that the protoplasm shrinks in size:
(a) What is this phenomenon called?
(b) What will happen if the filament is replaced in distilled water?
Soln.(a) When a freshly collected Spirogyra filament is kept in a 10% potassium nitrate solution the protoplasm shrinks in size, because the concentration of the solution is higher than the cell cytoplasm. This phenomenon is called plasmolysis.
(b) If the filament is replaced in distilled water it reabsorbs water and protoplast will come back in its original shape. This phenomenon is known as deplasmolysis.

14. Sugar crystals do not dissolve easily in ice cold water. Explain.
Soln. At low temperature, kinetic energy of the system decreases with more water molecules forming lattice structure. Few water molecules remain mobile. As a result water molecules striking sugar crystals are small in number. This results in slow dissolution of sugar crystals in ice cold water.

15.Salt is applied to tennis lawns to kill weeds. How does salting tennis lawns help in killing of weeds without affecting the grass?
Soln. Salting causes increase in the ion concentration outside the cells, it causes exosmosis from cell. As a result, cell cytoplasm shrinks and death of the plant will occur. Salting has no selective effects. It will kill all the plants on which it is sprayed. Yet, to prevent grasses present from being killed the following steps can be taken:
(i) Salt is sprayed selectively only on the weeds preventing it from affecting grass.
(ii)Grass plant are sprayed with water while weed plants are not. It helps in leaching salt down the soil, without allowing it sufficient time to contact with the grass plant and affect it.

16.What is the chemical composition of xylem and phloem sap?
Soln. Xylem sap consists of water and dilute concentration of most minerals.
Phloem sap consists of sucrose (5—10%), amino acids (1%), traces of bound minerals and water.

17. If you are provided with two tubes (A and B), where one is narrow and the other is relatively wider and if both are immersed in a beaker containing water as shown in the figure given.
ncert-exemplar-class-11-biology-solutions-transport-in-plants-5
Why does B show higher water rise than A?
Soln. Rise in water in the tubes is due to surface tension. Surface tension is inversely proportional to surface area. Surface area is less in narrow tube which, therefore has more surface tension and hence shows higher rise in water level.

18 .What are ‘aquaporins’? How does presence of aquaporins affect osmosis?
Soln. Aquaporins are membrane proteins which form pores or channels in membrane. It helps in transport of water soluble substances. In presence of aquaporins, osmosis becomes more rapid.

19. ABA (abscisic acid) is called a stress hormone.
(a) How does this hormone overcome stress conditions?
(b) From where does this hormone get released in leaves?
Soln.(a) Abscisic acid (ABA) is called stress hormone because it helps in overcoming stress condition by:
(i)Causing closure of stomata and reducing transpiration.
(ii)Reducing growth activity and metabolism.
(iii)inducing dormancy of buds and seeds, (b) Abscisic acid is released by mesophyll cells where it is formed from violaxantin.

20. We know that plants are harmed by excess water. But plants survive under flooded condition. How are they able to manage excess water?
Soln. Plant roots can tolerate temporary anaerobic conditions but not permanent anaerobiosis. Excess water in soil drives away air and causes permanent anaerobiosis. Flooding is a temporary phase. A lot of air remains trapped in the soil. Therefore, plants are able to tolerate flooding but not excess water or water logged conditions of soil as a permanent condition.

21.Differentiate between diffusion and translocation in plants.
Soln.Diffusion is the passage of substances from the region of their higher concentration to the region of their low concentration due to kinetic energy of the particles. It usually occurs in all directions.
Translocation is bulk transport of materials in solutions form, inside plant channels in particular directions caused by forces other than kinetic energy of the particles.

22.How is facilitated diffusion different from diffusion?
Soln. Facilitated diffusion is different from diffusion in following ways:
(i)Facilitated diffusion is passage of particles from the area of higher concentration to the area of lower concentration through specific sites (channel proteins) whereas diffusion is a passage of particles from the area of higher concentration to the area of lower concentration, throughout the available space.
(ii)In facilitated diffusion, only one type of particle can pass through its specific sites whereas in diffusion a number of different types of particles can diffuse simultaneously through the same space.
(iii)Facilitated diffusion stops or, slows down when saturation of carrier molecules occurs. In diffusion no saturation occurs.

23.Explain the mass flow hypothesis of transport in phloem.
Soln.Mass flow hypothesis was proposed by E. Munch (1930) and was elaborated by Crafts (1938). According to this hypothesis, organic substances move from the region of high osmotic pressure to the region of low osmotic pressure in a mass flow due to the development of a gradient of turgor pressure. Sieve tube system is fully adapted to mass flow of solutes. Here, the vacuoles are fully permeable because of the absence of tonoplast (Esau, 1966). A continuous high osmotic concentration is present in the source or supply region, e.g., mesophyll cells (due to photosynthesis). The organic substances present in them are passed into the sieve tubes through their companion cells by an active process. A high osmotic concentration, therefore, develops in the sieve tubes of the source. The sieve tubes absorb water from the surrounding xylem and develop a high turgor pressure. It causes the flow of organic solution towards the area of low turgor pressure. A low trugor pressure is maintained in the sink region by converting soluble organic substances into insoluble form. Water passes back into xylem.

24.0bserve the diagram and answer the following:
ncert-exemplar-class-11-biology-solutions-transport-in-plants-6
(a) Are these types of guard cells found in monocots or dicots?
(b) Which of these shows a higher water content (i) or (ii)?
(c) Which element plays an important role in the opening and closing of stomata?
Soln.(a) In given figure, guard cells are bean shaped which are mainly found in dicots.
(b) Figure (i) shows higher water content in the guard cells, because the cells are turgid and the stoma is open.
(c) K+ ions play an important role in the opening and closing of stomata.

25.Define uniport, symport and antiport. Do they require energy?
Soln. Uniport is one sided movement of a substance through the cell membrane independent of other solutes.Symport and antiport are two types of cotransport. Symport is simultaneous movement of two or more solutes across cell membrane in the same direction. Antiport is simultaneous movement of two solutes in opposite directions across the cell membrane.
ncert-exemplar-class-11-biology-solutions-transport-in-plants-7
Uniport, symport and antiport are types of facilitated diffusion which do not require energy. Similar type of passage also occur in case of active transport where energy is required.

Long Answer Type Questions
1. Minerals are present in the soil in sufficient amounts. Do plants need to adjust the types of solutes that reach the xylem? Which molecules help to adjust this? How do plants regulate the type and quantity of solutes that reach xylem?
Soln. Yes, plants need to adjust the types of solutes that reach the xylem. Transport proteins present on endodermis help to adjust this. The solute ions are checked and transported inwardly by transport proteins present over the endodermal cells. Endodermis allows the passage of ions inwardly but not outwardly. Endodermis acts as the checkpoint because before endodermis, water and dissolved minerals are carried upto the cortex by both apoplast and symplast. Symplastic movement can be regulated as plasma membrane is selectively permeable but apoplastic movement occurs through completely permeable non-living spaces, thus cannot be regulated. At the level
of endodermis, the apoplastic movement is stopped, due to presence of Casparian strip, hence only required minerals are allowed selectively to pass through the symplast. Plasma membranes of endodermal cells represented the final point at which the root could control entry of any disssolved solute.

2.Plants show temporary and permanent wilting. Differentiate between the two. Do any of them indicate the water status of the soil?
Soln. Wilting is the loss of turgidity of leaves and other soft aerial parts of a plant causing their drooping, folding and rolling. The symptoms of wilting are not shown by thick walled tissues. Therefore, they are less conspicuous in sclerophyllous plants.
It is mainly of three types:
(i) Incipient wilting
(ii) Temporary wilting
(iii)Permanent wilting
Incipient wilting shows no external symptoms of wilting.
Differences between temporary and permanent wilting are as follows:
(i) Temporary wilting is the temporary drooping down of leaves and young shoots due to loss of turgidity during noon. At this time the rate of transpiration is maximum. The rate of water absorption is less due to shrinkage of roots and depletion of water around the root hairs, lower leaves show wilting earlier than the upper ones. Whereas permanent wilting occurs when the soil is unable to meet the water requirement of plant.
(ii)Temporary wilting recovers as soon as water is replenished in the soil around root hair whereas permanent wilting cannot be recovered because cells do not regain their turgidity even in presence of plentiful water.
(iii)In temporary wilting plant regain its normal growth, whereas in permanent wilting plant eventually dies.

3.Which of these is a semipermeable membrane (S.P) and which is selectively permeable (S.L) (a) Animal bladder
(b) Plasmalemma
(c) Tonoplast
(d) Parchment membrane
(e) Egg membrane
Soln.
(a) Animal bladder – Selectively permeable membrane
(b) Plasmalemma – Selectively permeable membrane
(c) Tonoplast – Selectively permeable membrane
(d) Parchment membrane – Semipermeable mem¬brane
(e) Egg membrane – Selectively permeable membrane

4.Halophytes may show precell pressure very much higher than atmospheric pressure. Explain how this can happen?
Soln.Halophytes grow in soil with high concentration of salts. There is accumulation of salts in the cytoplasm, as a result the osmotic concentration of cytoplasm is increased, which causes entry of water into the cells. As a result, turgor pressure of the cells in the halophytes is comparatively higher. To limit it, halophytes perform two steps:
(i) They accumulate salts away from the cytoplasm in vacuoles.
(ii)They have special salt secreting glands which remove excess salts.

5.The radiolabelled carbon in carbon dioxide supplied to potato plants in an experiment was seen in the tuber eventually. Trace the movement of the labelled carbon dioxide.
Soln. If’ a radiolabelled carbon C14 is supplied to potato plants, they will carry out photosynthesis in the presence of light,
and theC14O2 will be fixed and will form radioactive products of photosynthesis, i.e., glucose C614 H12O6.
This radioactive glucose is converted to sucrose which would again be radioactive due to transfer of C14 from previous sugar molecule. These sucrose molecule then move into phloem and transported to other parts of plant and eventually reach the tuber where they are converted into radiolabelled starch and is stored.

6.Water molecule is very polar. Polar end of molecule attracts opposite charges on another water molecule (acts like magnet). How will you explain this property of water with reference to upward movement of water? Comment on the upward movement of Soln.water given the intermolecular hydrogen bonding in water.
Soln. Water molecules remain attached to one another by a strong mutual force of attraction called cohesion force, the mutual attraction is due to hydrogen bonds formed amongst adjacent water molecules.
On account of cohesion force, the water column can bear a tension or pull of up to 100 atm (Mac Dougal, 1936). Therefore, the cohesion force provides the tensile strength. Its theoretical value is about 15000 atm but the. measured value inside the tracheary elements ranges between 45 atm to 207 atm (Dixon and Jolly, 1894). Water column does not further break its connection from the tracheary elements (vessels and tracheids) because of another force called adhesion force between their walls and water molecules. Water molecules in liquid state are attracted to one another more than the water molecules in the gaseous state. It produces surface tension that accounts for high capillarity through tracheids and vessels.

7.Comment on the experimental setup.
ncert-exemplar-class-11-biology-solutions-transport-in-plants-8
(a) What does the setup demonstrate?
(b) What will happen to the level of water if a blower is placed close to setup.
(c) Will the mercury level fluctuate (go up/ down) if phenyl mercuric acetate is sprayed on leaves?
Soln.(a) The given setup is demonstration of pull due to transpiration.
(b) If a blower is placed close to set up, rate of transpiration will increase because movement of air increases the rate of transpiration. The air from blower is low in moisture concentration and thus will
increase transpiration. Now, humid air will move away quickly (as the blower is still on) and the newly arrived dry air will cause further transpiration. As a result, overall rate of transpiration will increase. If rate of transpiration increases, water level will go down.
(c) Phenyl mercuric acetate is an antitranspirant. If it is sprayed on leaves, transpirational loss of water will stop and the mercury level will remain stable.

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NCERT Exemplar Class 11 Biology Solutions Biomolecules

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NCERT Exemplar Class 11 Biology Solutions Biomolecules

Multiple Choke Questions
1.It is said that elemental composition of living organisms and that of inanimate objects (like earth’s crust) are similar in the sense that all the major elements are present in both. Then what would be the difference between these two groups?
Choose a correct answer from the following.
(a) Living organisms have more gold in them than inanimate objects.
(b) Living organisms have more water in their body than inanimate objects.
(c) Living organisms have more carbon, oxygen and hydrogen per unit mass than inanimate objects.
(d) Living organisms have more calcium in them than inanimate objects.
Soln.(c): A comparison of elements present in non-living and living matter is given in the table below:
ncert-exemplar-class-11-biology-solutions-biomolecules-1

2.Many elements are found in living organisms
either free or in the form of compounds. One of the following is not found in living organisms,
(a) Silicon (b) Magnesium
(c) Iron (d) Sodium
Soln.(a): Various mineral elements present in living organisms include (i) major elements carbon (C), nitrogen (N), oxygen (O), hydrogen (H), phosphorus (P), calcium (Ca), magnesium (Mg), sodium (Na), potassium (K) and chlorine (Cl) and (ii) trace elements, iodine (I), iron (Fe), manganase (Mn), cobalt (Co), copper (Cu), zinc (Zn), molybdenum (Mo), boron etc.

More Resources for CBSE Class 11

3.Amino acids have both an amino group and a carboxylic acid group in their structure. Which amongst the following is an amino acid?
(a) Formic acid (b) Glycerol
(c) Glycolic acid (d) Glycine
Soln.(d): Glycine is the simplest amino acid with hydrogen as R group. Formic acid is simplest carboxylic acid molecule. It is chemically HCOOH. Glycerol is a polyol compound, which is chemically propane-1, 2, 3-triol and glycolic acid is chemically, 2-hydroxyethanoic acid.

4.An amino acid under certain conditions have both positive and negative charges simultaneously in the same molecule. Such a form of amino acid is called
(a) acidic form
(b) ba’sicform
(c) aromatic form
(d) zwitterionicform.
Soln.(d) :
ncert-exemplar-class-11-biology-solutions-biomolecules-2
In neutral solution, an amino acid molecule exists as a dipolar ion (zwitter ion) i.e., molecule containing positive and negative ionic groups. The charge on the ion changes with pH. In an acidic solution (low pH) the amino acid picks upH+ ions and becomes positively charged. In alkaline solution (high pH) the amino acid donates H+ ions to the medium and becomes negatively charged. The pH at which the amino acid is electrically neutral or the molecule exists as a zwitterion is called the isoelectric pH.

5.Which of the following sugars have the same number of carbon as present in glucose?
(a) Fructose (b) Erythrose
(c) Ribulose (d) Ribose
Soln. (a): Fructose and glucose, both are
six carbon sugars having formula C6 H12 O6 Erythrose is a four carbon sugar. Ribulose and ribose are both five carbon sugars.

6.An acid soluble compound formed by
phosphorylation of nucleoside is called
(a) nitrogen base
(b) adenine
(c) sugar phosphate
(d) nucleotide.
Soln.(d) : Nucleoside is basically pentose sugar + nitrogenous base. Nucleotide is pentose sugar + nitrogenous base + phosphate group. DNA and RNA are composed of nucleotides.

7.When we homogenise any tissue in an acid the acid soluble pool represents
(a) cytoplasm (b) cell membrane
(c) nucleus (d) mitochondria.
Soln.(a): The acid soluble pool has roughly similar composition as that of cytoplasm. Biomolecules with molecular weights in the range of 18 – 800 Daltons come in acid- soluble fraction (with the exception of lipids). Though, the macromolecules from cytoplasm and organelles represent the acid-insoluble fraction.

8.The most abundant chemical in living organ-isms could be
(a) protein (b) water
(c) sugar (d) nucleic acid.
Soln.(b)

9.A homopolymer has only one type of building block called monomer repeated ‘n’ number of times. A heteropolymer has more than one type of monomer. Proteins are heteropolymers usually made of
(a) 20 types of monomers
(b) 40 types of monomers
(c) 30 types of monomers
(d) only one type of monomer.
Soln.(a): The function and shape of a protein is affected by sequence of 20 types of amino acids, each having an amino group -NH2, a carboxylic acid group -COOH, a hydrogen atom each attached to carbon located next to -COOH group and a side chain R which varies from one amino acid to other. (It can be hydrogen or an aliphatic group, an aromatic or heterocyclic group).
ncert-exemplar-class-11-biology-solutions-biomolecules-3

10.Proteins perform many physiological functions. For example, some proteins function as enzymes. One of the following represents an additional function that some proteins performs.
(a) Antibiotics
(b) Pigment conferring colour to skin
(c) Pigment making colours of flowers
(d) Hormones.
Soln.(d) :Most of the body functions are regulated by hormones like growth, vegetative and sexual development, thermal regulation, cellular oxidation, metabolism of carbohydrates, proteins, fats etc. Hormones are needed in very small quantity to carry out these functions. Some hormones are proteinaceous, e.g., insulin (regulates sugar metabolism), growth hormone of pituitary, parathyroid hormone i.e., parathormone (regulates Ca and phosphate transport).

11.Glycogen is a homopolymer made up of (a) glucose units (b) galactose units (c) ribose units (d) amino acids.
Soln.(a): Glycogen is a branched polymer of glucose. It is readily soluble in water. It consists of a-D glucose units, mostly linked by 1-4, a glycosidic linkage, and is highly branched via frequent 1-6 linkages. Glycogen is found mostly in muscles and liver of animals and is also called animal starch. It gives red colour with iodine solution. It has about 30,000 glucose residues and a molecular weight of about 4.8 million. The straight part is helically twisted with each turn having six glucose units. The distance between two branching points is 10-14 glucose residues. In a polysaccharide chain of glycogen, the right end is called reducing end and the left end is called non-reducing end.

12. The number of ends’ in a glycogen molecule would be
(a) equal to the number of branches plus one
(b) equal to the number of branch points
(c) one
(d) two, one on the left side and another on the right side.
Soln.(a)

13. The primary structure of a protein molecule has
(a) two ends   (b) one end
(c) three ends (d) no ends.
Soln.(a): Primary structure of a protein is simply the amino acid sequence of it which has two ends the carboxyl and amino terminals.

14. Which of the following reactions is not enzyme- mediated in biological system?
(a) Dissolving C02 in water
(b) Unwinding the two strands of DNA
(c) Hydrolysis of sucrose
(d) Formation of peptide bond
Soln.(a) : DissolvingC02 in water does not require any enzyme. C02 has higher solubility in water than 02. Solubility of C02 in water can be increased with decrease in temperature, a principle used in carbonated drinks.

Very Short Answer Type Questions
1.Medicines are either man made (i.e., synthetic) or obtained from living organisms like plants, bacteria, animals etc. and hence the latter are called natural products. Sometimes natural products are chemically altered by man to reduce toxicity or side effects. Write against each of the following whether they were initially obtained as a natural product or as a synthetic chemical.
(a) Penicillin________
(b) Sulfonamide________
(c) Vitamin C________
(d) Growth hormone_______
Soln.
(a) Natural
(b) Synthetic
(c) Natural
(d) Natural

2.Select an appropriate chemical bond among ester bond, glycosidic bond, peptide bond and hydrogen bond and write against each of the following.
(a) Polysaccharide
(b) Protein
(c) Fat
(d) Water
Soln. (a) Glycosidic bond
(b) Peptide bond
(c) Ester bond
(d) Hydrogen bond

3.Write the name of any one aminoacid, sugar, nucleotide and fatty acid.
Soln. Amino acid – Lysine Sugar – Fructose
Nucleotide – Adenosine monophosphate Fatty acid – Palmitic acid

4.Reaction given below is catalysed by
Soln.oxidoreductase between two substrates A and A’, complete the reaction. A reduced + A’ oxidised >
HDOxidoreductase is an enzyme that takes part in oxidation and reduction reactions. This enzyjne is involved in transfer of electrons from one molecule called as electron donor to another molecule called electron acceptor. The complete reaction is :
ncert-exemplar-class-11-biology-solutions-biomolecules-4

5.How are prosthetic groups different from co-factors?
Soln. Co-factors are the non-protein constituents of conjugated or compound enzymes, associated with the apoenzyme of the enzyme (protein portion). Co-factor is small, heat-stable and dialysable part of conjugate enzyme. It may be inorganic, e.g., Ca, Fe, Cu, Zn etc. or organic (e.g. prosthetic group and coenzyme). Prosthetic groups are non-protein organic cofactors firmly attached to the apoenzymes, e.g., haem, biotin, etc.

6.Glycine and alanine are different with respect to one substituent on the a-carbon. What are the other common substituent groups?
Soln. General formula of amino acid is —
ncert-exemplar-class-11-biology-solutions-biomolecules-5
In Glycine, R is represented by hydrogen (H) and in alanine, by methyl group(CH3).  COOH,H2N and H are common substituent groups in both glycine and alanine.

7.Starch, cellulose, glycogen, chitin are polysac-charides found among the following. Choose the one appropriate and write against each.
Cotton fibre________
Exoskeleton of cockroach_________
Liver________
Peeled potato________
Soln.
Cotton fibre – cellulose
Exoskeleton of cockroach – chitin
Liver – glycogen
Peeled potato – starch

Short Answer Type Questions
1.Enzymes are proteins. Proteins are long chains of aminoacids linked to each other by peptide bonds. Aminoacids have many functional groups*in their structure. These functional groups are, many of them at least, ionisable. As they are weak acids and bases in chemical nature, this ionization is influenced by pH of the solution. For many enzymes, activity is influenced by surrounding pH. This is depicted in the curve below, explain briefly.
ncert-exemplar-class-11-biology-solutions-biomolecules-6
Soln. Everv enzyme has an optimum pH when it is most effective. A rise or fall in pH reduces enzyme activity by changing the degree of ionisation of its side chains. A change in pH may also start reverse reaction. Fumarase catalyses  fumarate -> malate at 6.2 pH and reverse at 7.5 pH. Most of the intracellular enzymes function near neutral pH with the exception of several digestive enzymes which work either in acidic range of pH or alkaline, e.g., 2 pH for pepsin and 8.5 pH for trypsin.

2.Is rubber a primary metabolite or a secondary . metabolite? Write four sentences about rubber.
Soln. Rubber is a secondary metabolite. Natural rubber is a polymer of 2 methyl-1, 3-butadiene which is more commonly known as isoprene. Its molecular weight is typically between 100,000 and a million which is equivalent to between 1500 and 15000 monomer units. Rubber is extracted from latex which is a milky suspension produced by plants to heal wounds. Latex has rubber, water, resins, terpenes, proteins and sugars. Latex is collected from rubber trees using a process called tapping. In natural state rubber deforms easily but is not very elastic because the long, unbranched polymer can easily slide past each other. It is also very tacky, which makes it useful for application such as adhesives. However, for a majority of its uses rubber is crosslinked using sulphur in a process called vulcanisation.

3.Schematically represent primary, secondary and tertiary structures of a hypothetical polymer say for example a protein.
Soln.Primary structure of a protein includes number of polypeptides, number and sequence of amino acids in each polypeptide.
ncert-exemplar-class-11-biology-solutions-biomolecules-7
The development of new stearic relationships of amino acids present in linear sequence inside the polypeptides leads to the formation of secondary structure of proteins.
ncert-exemplar-class-11-biology-solutions-biomolecules-8
Tertiary structure involves interactions that are caused by binding and folding of a-helix or P sheets leading to the formation of rods, spheres or fibres. Tertiary structure is stabilised by several types of bonds such as, H-bonds, ionic bonds, covalent bonds, van der Waal’s interactions hydrophobic bonds, etc.
ncert-exemplar-class-11-biology-solutions-biomolecules-9

4.Nucleic acids exhibit secondary structure, justify with example.
Soln. Nucleic acids show a wide variety of secondary structures. DNA or deoxyribose
nucleic acid is a helically twisted double chain polydeoxyribonucleotide macromolecule. A DNA molecule has two unbranched comple¬mentary strands. They are spirally coiled. The two spiral strands of DNA are collectively called DNA duplex. The two strands are not coiled upon each other but double strand is coiled upon itself around common axis. The double helical structure of DNA is stabilised by phosphodiester bond, hydrogen bond and ionic interactions.
5. Comment on the statement “living state is a non-equilibrium steady state to be able to perform work”.
Soln.Metabolites or biomolecules occur in organisms in concentrations characteristic of each of them. The living systems maintain this concentration of biomolecules because they are in metabolic flux, always remaining in nonequilibrium steady state where equilibrium is seldom achieved. No work can be carried out in equilibrium state. Therefore, living systems are regularly receiving an input of energy to prevent reaching an equilibrium and remain always in nonequilibrium steady state. Energy is obtained from metabolism. Metabolism and living state are, therefore, complementary. There cannot be a living state without metabolism.

Long Answer Type Questions
1.Formation of enzyme-substrate complex (ES) is the first step in catalysed reactions. Describe the other steps till the formation of product.
Soln.Each enzyme has an active site. The active sites of enzymes have a specific conformation for attracting and holding substrate. Both enzyme and substrate molecules have specific geometrical shapes. In the region of active sites the surface configuration of the enzyme is such as to allow the particular substrate molecules to be held over it. The contact is such that the substrate molecules or reactants come together causing the chemical change. It is similar to the system of lock and key. Just as a lock can be opened by its specific key, a substrate molecule can be acted upon by a particular enzyme. After coming in contact with the active site of the enzyme, the substrate molecules or reactants form a complex called enzyme-substrate complex. The active site of enzyme is now in close proximity with the substrate and break its chemical bonds and a new enzyme product complex is formed. The products are soon time so that an enzyme-product complex is also formed. However, the products are soon released and the freed enzyme is able to bind more substrate molecules.
ncert-exemplar-class-11-biology-solutions-biomolecules-19
Thus we see that the chemical reactants do not cause any alteration in the composition or physiology of the enzyme. The same enzyme molecule can be used again and again. Hence, enzymes are required in very small concentrations.
ncert-exemplar-class-11-biology-solutions-biomolecules-10
Upper series – Breakdown reaction Lower series – Biosynthetic reaction

2.What are different classes of enzymes? Explain any two with the type of reaction they catalyse.
Soln. According to International Union of Biochemistry (IUB) enzymes are grouped into the following six categories.
(i) ’Oxidoreductases : They take part in oxidation and reduction reactions or transfer of electrons.
ncert-exemplar-class-11-biology-solutions-biomolecules-11
Oxidoreductases are of three types- oxidases, dehydrogenases and reductases, c.g., cyto-chrome oxidase, succinate dehydrogenase, nitrate reductase.
(ii) Transferases : They transfer a group from one molecule to another c.g., glutamate pyruvate transaminase (transfers amino group from glutamate to pyruvate during synthesis of alanine). The chemical group transfer does not occur in the free state.
ncert-exemplar-class-11-biology-solutions-biomolecules-12
(iii) Hydrolases : They catalyse hydrolysis of bonds like ester, ether, peptide, glycosidic, C—C, C—halide, P—N, etc. which are formed by dehydration condensation. Hydrolases break up large molecules into smaller ones with the help of hydrogen and hydroxyl groups of water molecules. The phenomenon is called hydrolysis. Digestive enzymes belong to this group, e.g., amylase (hydrolysis of starch) sucrase, lactase.
ncert-exemplar-class-11-biology-solutions-biomolecules-13
(iv)Lyases : These enzymes cause cleavage, removal of groups without hydrolysis, addition of groups to double bonds or removal of a group producing double bond, e.g., histidine decarboxylase (breaks histidine to histamine and COz), aldolase (fructose-1, 6-diphosphate to dihydroxy acetone phosphate and glyceraldehyde phosphate).
ncert-exemplar-class-11-biology-solutions-biomolecules-14
(v)Isomerases : These enzymes cause rearrangement of molecular structure to effect isomeric changes. They are of three types, isomerases (aldose to ketose group or vice-versa like glucose 6-phosphate to fructose 6-phosphate), epimerases (change in position of ope constituent or carbon group like xylulose phosphate to ribulose phosphate) and mutases (shifting the position of side group like glucose-6-phosphate to glucose-1- phosphate).
ncert-exemplar-class-11-biology-solutions-biomolecules-15
(vi)Ligases (Synthetases) : These enzymes catalyse bonding of two chemicals with the help of energy obtained from ATP resulting in formation of such bonds as C—O, C—S, C—N and P—O, e.g., pyruvate carboxylase. It combines pyruvic acid with C02 to produce oxaloacetic acid.
Pyruvic acid + C02 + ATP + HzO
ncert-exemplar-class-11-biology-solutions-biomolecules-16

3.Nucleic acids exhibit secondary structure. Describe through Watson-Crick Model.
Soln.Nucleic acids are long chain
macromolecules which are formed by end to end polymerisation of large number of repeated units called nucleotides. Nucleic acids show a variety of secondary structures. There are two types of nucleic acids – deoxyribonucleic acid or DNA and ribonucleic acid or RNA. Deoxyribonucleic acid (DNA) is genetic material found in the nucleus of all living cells with the exception of some viruses. The structure of DNA was elucidated by Watson and Crick based on X-ray diffraction studies. They proposed a double helix model of DNA. According to this model, DNA exists as a double helix and consists of two strands of polynucleotides that are antiparallel to each other, i.e., both run in opposite directions, one in 5′ -> 3′ direction and other in 3′ -> 5′ direction.
The backbone of DNA strand is formed by alternate sugar and phosphoric acid group. The phosphate group is connected to carbon 5’of the sugar residues of its own nucleotide and 3′ of the sugar residue of the next nucleotide by phosphodiester bonds.The nitrogen bases are projected more or less perpendicular to the backbone of DNA and face inside. A and G of one strand base pair with T and C respectively on the other strand. Between A and T (A=T), there are two hydrogen bonds while, there are three hydrogen bonds between G and
ncert-exemplar-class-11-biology-solutions-biomolecules-21

4.What is the difference between a nucleotide and nucleoside? Give two examples of each with their structure.
Soln. Differences between nucleoside and nucleotide are as follows :
ncert-exemplar-class-11-biology-solutions-biomolecules-18

5.Describe various forms of lipid with a few examples.
Soln. Lipids are fatty acids esters of alcohols and related substances which are insoluble in water but get dissolved in a number of non-polar organic solvents like ether, benzene, chloroform, acetone, etc.
Depending upon their composition and characteristics lipids are often classified into . simple lipids, compound lipids and derived lipids.
(i) Simple lipids – These are formed from fatty acids and alcohol. They do not have any additional group, e.g., fats, suberin, cutin, wax.
(a) Neutral or true fats – They are triglycerides which are formed by esterification of three molecules of fatty acids with one molecule of trihydric alcohol, glycerol (glycerine or trihydroxy propane). Three molecules of water are eliminated.
In fats the three fatty acids are only rarely similar (e.g., tripalmitin, tristearin, triolein). They are called pure fats. Usually they are dissimilar or two of the three fatty acids are similar. They are known as mixed fats, e.g., Butter. Fats are named after the names of fatty acids, e.g., dipalmito-stearin, palmito-oleio-stearin, steario-oleio-palmitin.
(b) Waxes – They are fatty acid esters of long chain monohydric alcohols like cytyl, ceryl or mericyl.
(c) Cutin – It is a complex lipid produced by cross-esterification and polymerisation of hydroxy fatty acids, as well as other fatty acids with or without esterification by alcohols other than glycerol.
(d) Suberin – It is a mixture of fatty material having condensation products of glycerol and phellonic acid or its derivatives.
(ii)Compound or conjugated lipids: These are the esters of fatty acids and alcohol but contain other substances also, e.g., phospholipid, glycol’ipids, sphingolipids etc.
(a) Phospholipids- They are triglyceride
lipids where one fatty acid is replaced
by phosphoric acid which is often linked to ‘additional nitrogenous groups like choline (in lecithin), ethanolamine (in cephalin), serine or inositol. ”
(b) Sphingolipids – They are lipids
having amino alcohol sphirigosine.
Sphingomyelins contain an additional
phosphate attached to choline like
phospholipids.
(c) Glycolipids – These are sugar containing
lipids, in which the lipids portion of the
molecule is usually based on glycerol or sphingosine and the sugar is typically
galactose, glucose or inositol.
(iii)Derived lipids: These are lipid-like substance such as sterol or derivatives of lipids, e.g., steroids, prostaglandins and teapenes.
(a) Steroids – They are a group of complex
lipids that possess a hydrogenated
cyclopentano-perhydrophenanthrene
ring system.
(b) Prostaglandins – They are derivatives
of arachidonic acid and other 20 carbon
fatty acids.
(c) Terpenes – They are lipid like hydrocarbons
formed of isoprene (C5H8) units. Steroids like cholesterol are also derived from terpenes having 6 isoprene units. .
Fats are also differentiated into two main
types, on the basis of their melting points at room temperature as follows.
(a) Hard Fats are solids at room temperature and contains long chains of fatty acids, e.g., Animals fat.
(b) Oils are usually liquid at room temperature
because they have low melting point, e.g.,
groundnut (peanut) oil, cotton seed oil, mustard oil, etc.

The post NCERT Exemplar Class 11 Biology Solutions Biomolecules appeared first on Learn CBSE.

NCERT Solutions for Class 7th Maths Chapter 6 The Triangle and its Properties Exercise 6.4


NCERT Solutions for Class 7th Maths Chapter 8 Comparing Quantities Exercise 8.2

NCERT Solutions for Class 7th Maths Chapter 8 Comparing Quantities Exercise 8.3

NCERT Solutions for Class 7th Maths Chapter 11 Perimeter and Area Exercise 11.2

NCERT Solutions for Class 7th Maths Chapter 11 Perimeter and Area Exercise 11.3

NCERT Solutions for Class 7th Maths Chapter 11 Perimeter and Area Exercise 11.4

NCERT Solutions for Class 7th Maths Chapter 12 Algebraic Expressions Exercise 12.1

NCERT Solutions for Class 7th Maths Chapter 12 Algebraic Expressions Exercise 12.2


NCERT Solutions for Class 7th Maths Chapter 12 Algebraic Expressions Exercise 12.3

NCERT Solutions for Class 7th Maths Chapter 12 Algebraic Expressions Exercise 12.4

Chemistry Practicals Class 12 Viva Questions with Answers

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Chemistry Practicals Class 12 Viva Questions with Answers

Question.1. Why is a Bunsen burner provided with air holes?
Answer. To regulate the supply of air.

Question.2. What type of flame would you use for general heating purpose?
Answer. A non-luminous oxidising flame as it gives maximum heat due to complete combustion of hydrocarbons.

Question. 3. What is the use of a fume cup-board?
Answer. It is used to perform those experiments which involve the production of poisonous gases or vapours.

Question.4. What is the use of wash bottle?
Answer. It is used for getting a thin stream of water required for washing or transferring a precipitate.

Question. 5. What first aid is necessary when an acid gets into an eye while working in the laboratory?
Answer. The injured eye should be washed freely with water and then 1% solution of sodium bicarbonate.

Chemistry Lab ManualNCERT Solutions Class 12 Chemistry Sample Papers

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Determine the Equivalent Mass and Number of Molecules of Water of Crystallisation in a Sample of Mohr’s salt

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Determine the Equivalent Mass and Number of Molecules of Water of Crystallisation in a Sample of Mohr’s salt, FeSO4(NH4)2 SO4 . nH20. Provided KMnO4

Chemistry Lab ManualNCERT Solutions Class 12 Chemistry Sample Papers

Chemical Equations
determine-the-equivalent-mass-and-number-of-molecules-of-water-of-crystallisation-in-a-sample-of-mohrs-salt-1

Theory
Prepare a solution of Mohr’s salt with known strength (g/litre). Molarity of ferrous ammonium sulphate can be determined by directly titrating it against standard (M/100) KMnO4 solution.
Molecular mass = strength/molarity.
Substituting the value of strength and value of molarity as calculated above, the molecular mass of Mohr’s salt can be calculated. Suppose it comes out to be M.
Theoretical molecular mass of Mohr’s salt, FeSO4.(NH4)2SO4. nH20
= 284 + 18n
∴ 284 + 18n = M
whence, n can be calculated.
In case of Mohr’s salt equivalent mass is equal to its molecular mass. Therefore, Equivalent mass of Mohr’s salt = M.

Indicator
KMnO4 is a self-indicator.

End Point
Colourless to permanent pink colour (KMnO4 in burette).

Procedure

  1. Weigh exactly 4.90 g of Mohr’s salt and dissolve in water to prepare exactly 250 ml of solution, using a 250 ml measuring flask. Rinse the pipette with the prepared Mohr’s
    salt solution and pipette out 20.0 ml of it in a washed titration flask.
  2. Rinse and fill the burette with M/100 KMnO4 solution.
  3. Add one test-tube (~ 20 ml) full of dilute sulphuric acid (~ 4 M) to the solution in titration flask.
  4. Note the initial reading of the burette.
  5. Now add KMnO4 solution from the burette till a permanent light pink colour is imparted to the solution in the titration flask on addition of last single drop of KMnO4 solution.
  6. Note the final reading of the burette.
  7. Repeat the above steps 4-5 times to get a set of three concordant readings.

Observations
Weight of watch glass =……. g
Weight of watch glass + Mohr’s salt =…………..g
Weight of Mohr’s salt = 4.90 g
Volume of Mohr’s salt solution prepared = 250 ml
Volume of Mohr’s salt solution taken for each titration = 20.0 ml
Molarity of KMnO4 solution =M/100
determine-the-equivalent-mass-and-number-of-molecules-of-water-of-crystallisation-in-a-sample-of-mohrs-salt-2

Calculations
determine-the-equivalent-mass-and-number-of-molecules-of-water-of-crystallisation-in-a-sample-of-mohrs-salt-3
determine-the-equivalent-mass-and-number-of-molecules-of-water-of-crystallisation-in-a-sample-of-mohrs-salt-4

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Prepare 250 ml of a N/10 Solution of Oxalic Acid from Crystalline Oxalic Acid

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Prepare 250 ml of a N/10 Solution of Oxalic Acid from Crystalline Oxalic Acid

Chemistry Lab ManualNCERT Solutions Class 12 Chemistry Sample Papers

Theory
prepare-250-ml-of-a-n10-solution-of-oxalic-acid-from-crystalline-oxalic-acid-1
prepare-250-ml-of-a-n10-solution-of-oxalic-acid-from-crystalline-oxalic-acid-2

Apparatus
Watch glass, analytical balance, weight box, fractional weight box, 250 ml beaker, glass rod, 250 ml measuring flask and wash bottle.

Chemical Required
Oxalic acid crystals and distilled water.

Procedure

  1. Take a watch glass, wash it with distilled water and then dry it.
  2. Weigh the clean and dried watch glass accurately and record its weight in the note book.
    prepare-250-ml-of-m10-solution-of-oxalic-acid-from-crystalline-oxalic-acid-1
  3. Weigh 3.150 g oxalic acid on the watch glass accurately and record this weight in the note-book.
  4. Transfer gently and carefully the oxalic acid from the watch glass into a clean 250 ml beaker. Wash the watch glass with distilled water with the help of a wash bottle to transfer the particles sticking to it into the beaker [Fig].
    The volume of distilled water for this purpose should not be more than 50 ml.
  5. Dissolve oxalic acid crystals in the beaker by gentle stirring with a clean glass rod.
  6. When the oxalic acid in the beaker is completely dissolved, transfer carefully the entire solution from the beaker into a 250 ml measuring flask (volumetric flask) with the help of a funnel [Fig].
    prepare-250-ml-of-m10-solution-of-oxalic-acid-from-crystalline-oxalic-acid-2
  7. Wash the beaker with distilled water. Transfer the washings into the measuring flask [Fig].
  8. Finally wash the funnel well with distilled water with the help of a wash bottle to transfer the solution sticking to the funnel into the measuring flask [Fig].
  9. Add enough distilled water to the measuring flask carefully, up to just below the etched mark on it, with the help of a wash bottle.
  10. Add the last few drops of distilled water with a pipette until the lower level of the meniscus just touches the mark on the measuring flask [Fig].
  11. Stopper the measuring flask and shake gently to make the solution uniform through-out. Label it as oxalic acid solution.
    prepare-250-ml-of-m10-solution-of-oxalic-acid-from-crystalline-oxalic-acid-3

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