Question Set A

CBSE 12th Standard Chemistry Electrochemistry HOT Questions Study Material - QB365

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Questions:

Generic 5

3 x 5 = 15

In the button cells widely used in watches and other devices the following reaction takes place:
Zn(s) + Ag₂O(s) + H2O(l) \(\rightarrow\) Zn²⁺(aq) + 2Ag(s) + 2OH^–(aq)
Determine \(\Delta_{r} G^{\ominus} \text { and } E^{\ominus}\) for the reaction.

The half-life for decay of radioactive ¹⁴C is 5730 years. An archaeological artefact containing wood had only 80% of ¹⁴ C activity as found in a living tree. Calculate the age of the artefact

100 g of liquid A (molar mass 140 g mol ^-1) was dissolved in 1000 g of liquid B (molar mass 180 g mol^-1). The vapour pressure of pure liquid B was found to be 500 torr. Calculate the vapour pressure of pure liquid A and its vapour pressure in the solution if the total vapour pressure of the solution is 475 torr.

Generic 2

5 x 2 = 10

The decomposition of dimethyl ether leads to the formation of CH₄, H₂ and CO and the reaction rate is given by Rate = k[PCH₃OCH₃] ^3/2 . If the pressure is measured in bar and time in minutes then what are the units of the rate and rate constants ?

For the reaction R \(\longrightarrow\) P, the concentration of a reactant changes from 0.03M to 0.02M in 25 minutes. Calculate the average rate of reaction using units of time both in minutes and seconds.

Why cannot vitamin C be stored in our body?

The conductivity of 0.001028 mol L^-1 acetic acid is \(4.95\times { 10 }^{ -5 }S{ cm }^{ -1 }\) Calculate its dissociation constant if \({ \Lambda }_{ m }^{ o }\) for acetic os 390.5 S cm²mol^-1.

19.5 g of CH₂FCOOH is dissolved in 500 g of water. The depression in the freezing point of water observed is 1.0^oC. Calculate the van't Hoff factor and dissociation constant of fluoro acetic acid.

Multiple Choice Question

18 x 1 = 18

The following experimental rate data were obtained for a reaction carried out at 25C
A(g)+ B(g) → C(g)+ D(g)

What are the orders with respect to A(g) and B(g)?

(a)

Order with respect to A(g)	Order with respect to B(g)
Zero	Second

(b)

Order with respect to A(g)	Order with respect to B(g)
First	Zero

(c)

Order with respect to A(g)	Order with respect to B(g)
Second	Zero

(d)

Order with respect to A(g)	Order with respect to B(g)
Second	First

10.

On the basis of following E^ovalues, the strongest oxidising agent is
[Fe(CN)₄)^4- \(\longrightarrow\) [Fe(CN)₆]^3- + e^-E^o = - 0.35 V
Fe²⁺ \(\longrightarrow\) Fe³⁺ + e^- E^o = - 0.77 V

(a)

Fe²⁺

(b)

Fe³⁺

(c)

[Fe(CN)₆]^3-

(d)

[Fe(CN)₆]^4-

11.

Which of the following expression is correct for 'Ka' in terms of A0 and A, where 'C' is molarity.

(a)

\(\mathbf{K}_{a}=\frac{\mathbf{C} \Lambda_{m}^{\circ}}{\Lambda_{m}\left(\Lambda_{m}^{\circ}-\Lambda\right)}\)

(b)

\(\mathbf{K}_{a}=\frac{\mathbf{C} \Lambda_{m}^{2}}{\Lambda_{m}^{0}\left(\Lambda_{m}^{0}-\Lambda_{m}\right)}\)

(c)

\(\mathbf{K}_{a}=\frac{\mathbf{C} \Lambda_{m}^{2}}{\Lambda_{m}^{\circ}}\)

(d)

\(\mathbf{K}_{a}=\frac{\mathbf{C} \Lambda_{m}^{2}}{\left(\Lambda_{m}^{0}-\Lambda_{m}\right)}\)

12.

K_b (molal elevation constant) is inversely proportional to

(a)

boiling point of solvent

(b)

\(\Delta_{\text {vap }}\)H of solvent

(c)

Molar mas of solvent

(d)

all of these

13.

If conc. of reactant' A' is increased 10 times and rate of reaction becomes 100 times. What is order with respect to 'A'?

(a)

(b)

(c)

(d)

14.

In the given reaction,
2Cu⁺(aq) \(\rightleftharpoons\) Cu²⁺(aq) + Cu(s)
E^O_Cu+_/Cu = 0.6 V and E^oCu²⁺/Cu = 0.41 V
Find out the equilibrium constant.

(a)

2.76 x 10²

(b)

2.76 X 10⁴

(c)

2.76 X 10⁶

(d)

2.76 x 10⁸

15.

DNA and RNA contain four bases each. which of the following bases is not present in RNA ?

(a)

Adenine

(b)

Uracil

(c)

Thymine

(d)

Cytosine

16.

Nucleic acids are the polymers of ......... .

(a)

Nucleosides

(b)

Nucleotides

(c)

Bases

(d)

Sugars

17.

Glycogen is a branched chain polymer of \(\alpha\)-D-glucose units nin which chain is formed by C₁ - C₄glycosidic linkage whereas branching occures by the formation of C₁-C₆ glycosidic linkage. Structure of clycogen is similar to ......... .

(a)

Amylose

(b)

Amylopectin

(c)

Cellulose

(d)

Glucose

18.

Which of the following relationships are not correct ?

(a)

pH of solution in hydrogen electrode

\({ =\frac { Electrode \ potential }{ 0.0591 } }at \ 298 \ K\)

(b)

\({ E }_{ cell }=\frac { 0.0591 }{ n } log{ K }_{ c }\)

(c)

Cell constant = Conductivity/Conductance

(d)

\({ \Delta G }^{ 0 }=nF{ E }_{ cell }^{ 0 }\)

19.

In the study of inversion of surcose in presence of readings at times r0 , r1 and \({ r }_{ \infty }\) represent the polarimetric readings at times 0, t and \( \infty \) respectively, then at the 50% inversion, which of the following relationship will hold good ?

(a)

\({ r }_{t }={ r }_{0 }+{ r }_{ \infty }\)

(b)

\({ r }_{t }\frac {1}{2}({ r }_{0 }+{ r }_{ \infty })\)

(c)

\({ r }_{t }={ r }_{0 }-{ r }_{ \infty }\)

(d)

\({ r }_{t }\frac {1}{2}({ r }_{0 }-{ r }_{ \infty })\)

20.

t_1/4 can be taken as the time taken for the concentration of a reactant to drop to \(\frac {3}{4}\) of its initial value. If the rate constant for a first order reaction is K, then t_1/4 can be written as

(a)

0.10/K

(b)

0.29/K

(c)

0.69/K

(d)

0.75/K

21.

The colligative properties of a solution are

(a)

\(\alpha \ molality\)

(b)

\(\alpha \ \frac { 1 }{ molecular \ mass \ of \ the \ solute } \)

(c)

proportional to each other

(d)

independent of the nature of the solute, i.e., electrolyte or non - electrolyte

22.

At a certain Hill station, water boils at 96^oC. The amount of NaCI that should be added to one litre of water so that it boils at 100^oC will be (K_b for H₂O = 0.52 K/m)

(a)

[Co(NH₃)₆]CI₃

(b)

[Co(NH₃)₅CI]CI₂

(c)

[Co(NH₃)₄CI₂]CI

(d)

None of these

23.

A first order reaction has a half-life period of 34.65 seconds. Its rate constant is

(a)

2 \(\times 10\) ^-2 sec^-1

(b)

4 \(\times 10\) ^-4 sec^-1

(c)

20 sec^-1

(d)

2 \(\times 10\) ^-4 sec^-1

24.

The standard emf of a galvanic cell involving 3 moles of electrons in a redox reaction is 0.59 V. The equilibrium constant for the reaction of the cell is

(a)

10²⁵

(b)

10²⁰

(c)

10¹⁵

(d)

10³⁰

25.

The values of van't Hoff factors for KCI, NaCI and K₂SO₄, respectively, are ................. .

(a)

2,2 and 2

(b)

2,2 and 3

(c)

1,1, and 2

(d)

1,1 and 1

26.

In which mode of expression, the concentration of solution remains independent of temperature ?

(a)

Molarity

(b)

Normality

(c)

Formality

(d)

Molality

Generic 3

5 x 3 = 15

27.

The standard electrode potential for Daniell cell is 1.1V. Calculate the standard Gibbs energy for the reaction:
Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s)

28.

Amongst the following compounds, identify which are insoluble, partially soluble and highly soluble in water?
(i) Phenol
(ii) Toluene
(iii) Formic acid
(iv) Ethylene glycol
(v) Chloroform
(vi) Pentanol

29.

Define the following as related to proteins.
(i) Peptide linkage
(ii) Primary structure
(iii) Denaturation.

30.

The rate of the chemical reaction doubles for an Increase of 10 K from 298 K. Calculate E_a.

31.

Calculate \(\Lambda_{m}^{0}\) for CaCl₂and MgSO₄ from the data given in Table

Ion	\(\lambda^0\)/(S cm²mol^–1)	Ion	\(\lambda^0\)/(S cm²mol^–1)
H⁺	349.6	OH^–	199.1
Na⁺	50.1	Cl^–	76.3
K+	73.5	Br–	78.1
Ca²⁺	119.0	CH₃COO	-40.9
Mg²⁺	106.0	\(\mathrm{SO}_{4}^{2-}\)	160.0

Case Study Questions

3 x 4 = 12

32.

Monosaccharides containing an aldehyde group are called aldoses while those containing a keto group are called ketoses. All monosaccharides containing five and six carbon atoms have cyclic structures, furanose (five-membered) and pyranose (six-membered). During ring formation, C₁ aldoses and C₂ in ketoses become chiral and hence all these monosaccharides exist in two forms called the α-anomer and β-anomer while C₁ and C2 are called glycosidic or anomeric carbon atoms. In contrast, stereoisomers, which differ in configuration at any other chiral carbon other than the glycosidic carbon are called epimers. Two molecules of the same or different monosaccharides combine together through glycosidic linkage to form disaccharides
1.Which of the following compounds show furanose structures?
a) Mannose
b) Galactose
c) Fructose
2.What is the relation between two molecules?
(a) Enantiomers
(b) Epimers
(c) Functional groups
(d) Anomers
3. In disaccharides, the linkage connecting monosaccharide units is called
(a) Glycogen linkage
(b) Nucleoside linkage
(c) Glycosidic linkage
(d) Peptide linkage.

33.

At 298 K, the vapour pressure of pure benzene, C₆H₆ is 0.256 bar and the vapour pressure of pure toluene C₆H₅CH₃ is 0.0925 bar. Two mixtures were prepared as follows:
(i) 7.8 g of C₆H₆ + 9.2 g of toluene
(ii) 3.9 g of C₆H₆ + 13.8 g of toluene
The following questions are multiple-choice questions. Choose the most appropriate answer:
(i) The total vapour pressure (bar) of solution 1 is
(a) 0.128
(b) 0.174
(c) 0.198
(d) 0.258
(ii) Which of the given solutions has higher vapour pressure?
(a) I
(b) II
(c) Both have an equal vapour pressure
(d) Cannot be predicted
(iii) Mole fraction of benzene in vapour phase in solution 1 is
(a) 0.128
(b) 0.174
(c) 0.734
(d) 0.266
(iv) Solution I is an example of a/an
(a) ideal solution
(b) non-ideal solution with a positive deviation
(c) non-ideal solution with a negative deviation
(d) can’t be predicted

34.

Read the passage given below and answer the following questions:
Carbohydrates are polyhydroxy aldehydes and ketones and those compounds which on hydrolysis give such compounds are also carbohydrates. The carbohydrates which are not hydrolysed are called monosaccharides. Monosaccharides with aldehydic group are called aldose and those which free ketonic groups are called ketose. Carbohydrates are optically active. Number of optical isomers = 2ⁿ
Where n = number of asymmetric carbons. Carbohydrates are mainly synthesised by plants during photosynthesis.
The monosaccharides give the characteristic reactions of alcohols and carbonyl group (aldehydes and ketones). It has been found that these monosaccharides exist in the form of cyclic structures. In cyclization, the -OH groups (generally C₅ or C₄ in aldohexoses and C₅ or C₆ in ketohexoses) combine with the aldehyde or keto group. As a result, cyclic structures of five or six membered rings containing one oxygen atom are formed, e.g., glucose forms a ring structure. Glucose contains one aldehyde group, one 1^o alcoholic group and four 2^o alcoholic groups in its open chain structure.
The following questions are multiple choice questions. Choose the most appropriate answer:
(i) First member of ketos sugar is

(a) ketotriose

(b) ketotetrose

(c) ketopentose

(d) ketohexose

(ii) In CH₂OHCHOHCHOHCHOHCHOHCHO, the number of optical isomers will be

(a) 16

(b) 8

(c) 32

(d) 4

(iii) Some statements are given below:
1. Glucose is aldohexose.
2. Naturally occurring glucose is dextrorotatory.
3. Glucose contains three, chiral centres.
4. Glucose contains one 1^o alcoholic group and four 2^o alcoholic groups.
Among the above, correct statements are

(a) 1 and 2 only	(b) 3 and 4 only
(c) 1,2 and 4 only	(d) 1,2,3 and 4

(iv) Which of the following reactions of glucose can be explained only by its cyclic structure?

(a) Glucose forms cyanohydrin with HCN

(b) Glucose reacts with hydroxylamine to form an oxime

(c) Pentaacetate of glucose does not react with hydroxylamine

(d) Glucose is oxidised by nitric acid to gluconic acid .

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Answers Key:

Generic 5

3 x 5 = 15

\( \mathrm{Zn}_{(s)} \longrightarrow \mathrm{Zn}^{2+}+2 \mathrm{e}^{-} ; E^{\ominus}=0.76 \mathrm{~V}\)
\(\mathrm{Ag}_{2} \mathrm{O}_{(s)}+\mathrm{H}_{2} \mathrm{O}_{(t)}+2 \mathrm{e}^{-} \longrightarrow 2 \mathrm{Ag}_{(s)}+2 \mathrm{OH}^{-} ; E^{\ominus}=0.344 \mathrm{~V} \)
__________________________________________________
\(\mathrm{Zn}_{(s)}+\mathrm{Ag}_{2} \mathrm{O}_{(s)}+\mathrm{H}_{2} \mathrm{O}_{(n)} \longrightarrow \mathrm{Zn}^{2+}+2 \mathrm{Ag}_{(s)}+2 \mathrm{OH}^{-} ; E^{\ominus}=1.104 \mathrm{~V}\)
E^ø= 1.104 V
We know that,
ΔrG^ø = -nFE^ø
= - 2 x 96487 x 1.04
= - 213043.296 J
= - 213.04 kJ

\({ t }_{ { 1 }/{ 2 } }=5730 \ years\)
\({ t }_{ { 1 }/{ 2 } }=\frac { 0.693 }{ k } \)
\(k=\frac { 0.693 }{ 5730 } { years }^{ -1 }\)
\(k=\frac { 2.303 }{ t } \log { \frac { { \left[ N \right] }_{ 0 } }{ \left[ N \right] } }\)
\(t=\frac { 2.303 }{ k } \log { \frac { 100 }{ 80 } }\)
\(=\frac { 2.303 }{ 0.693 } \times 5730\left[ \log { 5 } -\log { 4 } \right] \)
\(=\frac { 5730 }{ 0.3010 } \times \left[ 0.6990-0.6021 \right] \)
\(=\frac { 5730 }{ 0.3010 } \times 0.0969=\frac { 55.237 }{ 0.3010 } \)
\(=1844.64 \ years\approx 1845 \ years\)

\(\text { No. of moles of solute, } n_{2}=\frac{100}{140}=\frac{5}{7} \text { mole }\)
\(\text { No. of moles of solvent, } n_{1}=\frac{1000}{180}=\frac{50}{9} \text { mole }\)
Mole fraction of solute
\(x_{2}=\frac{n_{2}}{n_{1}+n_{2}}=\frac{5 / 7}{5 / 7+50 / 9}=0.114\)
\(\text {Mole fraction of solvent, } x_{1}=\left(1-x_{2}\right)=(1-0 \cdot 114)\)
= 0.886
According to Raoult's law
\( P_{A}=x_{A} P_{A}^{0}=0.114 \times P_{A}^{0} \)
\(P_{B}=x_{B} P_{B}^{\circ}=0.886 \times 500=443 \text { torr } \)
\(P_{\text {Total }}=P_{A}+P_{B}\)
\(475=0.114 P_{A}^{\circ}+443\)
\( P_{A}^{\circ}=\frac{475-443}{0 \cdot 114}=280 \cdot 7 \text { torr } \)
\(\therefore P_{A}=0 \cdot 114 \times 280 \cdot 7=32 \text { torr. }\)

Generic 2

5 x 2 = 10

\(Rate=\frac { bar }{ min } =bar \ { min }^{ -1 }\)
\(Rate=k{ \left( { P }_{ { CH }_{ 3 }O{ CH }_{ 3 } } \right) }^{ { 3 }/{ 2 } }\)
\(bar \ { min }^{ -1 }=k{ \left( bar \right) }^{ { 3 }/{ 2 } }\)
\(\\ k=\frac { bar \ { min }^{ -1 } }{ { \left( bar \right) }^{ { 3 }/{ 2 } } } ={ bar }^{ -{ 1 }/{ 2 } }{ min }^{ -1 }\)

\(Rate=-\frac { \Delta \left[ R \right] }{ \Delta t } \)
\(=-\frac { \left[ 0.02M-0.03M \right] }{ 25 } \)
\(=\frac { 0.01 }{ 25 } mol\quad { L }^{ -1 }\quad { min }^{ -1 }\)
\(Rate=0.4\times { 10 }^{ -3 }mol\quad { L }^{ -1 }\quad { min }^{ -1 }\)
\(=4\times { 10 }^{ -4 }mol\quad { L }^{ -1 }\quad { min }^{ -1 }\)
\(=\frac { 4\times { 10 }^{ -4 }mol\quad { L }^{ -1 } }{ 60\quad s } \)
\(=0.066\times { 10 }^{ -4 }\)
\(=6.6\times { 10 }^{ -6 }mol\quad { L }^{ -1 }\quad { s }^{ -1 }\)

Vitamin C cannot be stored in our body because it is soluble in water therefore gets excreted in urine.

\({ \wedge }_{ m }=\frac { 1000K }{ C } =\frac { 1000{ cm }^{ 3 }\times 4.95\times { 10 }^{ -5 }S{ cm }^{ -1 } }{ 0.001028mol } =48.15S{ cm }^{ 2 }{ mol }^{ -1 }\)
\(\\ \alpha =\frac { { \wedge }_{ m } }{ { { \wedge }^{ ° } }_{ m } } =\frac { 48.15S{ cm }^{ 2 }{ mol }^{ -1 } }{ 390.5S{ cm }^{ 2 }{ mol }^{ -1 } } =0.1233\)
\({ K }_{ \alpha }=\frac { { C\alpha }^{ 2 } }{ 1-\alpha } =\frac { 0.001028 }{ 1-0.1233 }\) = 1.78 × 10^–5 mol L^–1

Van't Hoff factor (i) = \(\frac { ({ M }_{ 2 })_{ cal } }{ ({ M }_{ 2 })_{ obs } } =\frac { 78 }{ 72.54 } =1.0753.\)
Taking volume of the solution as 500mL,
\({ K }_{ \alpha }=\frac { { C\alpha }^{ 2 } }{ 1-\alpha } =\frac { (0.5){ (0.0753) }^{ 2 } }{ 1-0.0753 } =3.07\times { 10 }^{ -3 }\) .

Multiple Choice Question

18 x 1 = 18

10.

(b)
Fe³⁺

11.

(b)
\(\mathbf{K}_{a}=\frac{\mathbf{C} \Lambda_{m}^{2}}{\Lambda_{m}^{0}\left(\Lambda_{m}^{0}-\Lambda_{m}\right)}\)

12.

(b)
\(\Delta_{\text {vap }}\)H of solvent

13.

(b)
2

14.

15.

16.

(b)
Nucleotides

17.

(b)
Amylopectin

18.

(a)
pH of solution in hydrogen electrode

\({ =\frac { Electrode \ potential }{ 0.0591 } }at \ 298 \ K\)

19.

(b)
\({ r }_{t }\frac {1}{2}({ r }_{0 }+{ r }_{ \infty })\)

20.

(b)
0.29/K

21.

(a)
\(\alpha \ molality\)

22.

(a)
[Co(NH₃)₆]CI₃

23.

(a) : \(k = \frac {0.693}{t/2}= \frac {0.693}{34.65}s^{-1} = 2 \times 10^{-2}\)s^-1

24.

(d)
10³⁰

25.

(b)
2,2 and 3

26.

(d)
Molality

Generic 3

5 x 3 = 15

27.

\(\Delta_{\mathrm{r}} G^{\ominus}=-n F \mathrm{E}_{\text {(cell) }}^{\ominus}\)
n in the above equation is 2, F = 96487 C mol ^–1 and \(\mathrm{E}_{\text {(cell })}^{\ominus}\) = 1.1 V
Therefore, \(\Delta_{\mathrm{r}} G^{\ominus}\) = – 2 × 1.1V × 96487 C mol ^–1
= – 21227 J mol^–1
= – 212.27 kJ mol^–1

28.

(i) Phenol Partially soluble (Reason: Phenol has polar - OH group and non-polar - C₆H₅ group).
(ii) Toluene Insoluble (Reason: Toluene is non-polar, water is polar.)
(iii) Formic acid Highly soluble (Reason: Hydrogen bonding)
(iv) Ethylene glycol Highly soluble (Reason: Hydrogen bonding).
(v) Chloroform Insoluble (Reason: H-bonds are forme although polarity is present).
(vi) Pentanol Partially soluble (Reason: - OH group is polar but long hydrocarbon part is non-polar).

29.

(i) Peptide linkage. Peptide bond is formed by the condensation of two or more same or different n-amino acids. The condensation occurs between amino acids with the elimination of water. In this case, the carboxyl group of one amino acid and amino group of another amino acid get condensed with the elimination of water molecule.
The resulting \(\quad O\\ \quad \parallel \\ -C-NH-\) linkage is called peptide linkage. The formation of a dipeptide and the peptide

(ii) Primary structure. The primary structure of proteins gives the sequence in which the amino acids are linked in one or more polypeptide chains of proteins. This is shown below:

(iii) Denaturation. A process that changes the physical and biological properties of proteins without affecting the chemical composition of a protein is called denaturation. The denaturation is caused by certain physical or chemical treatments such as changes in pH, temperature, presence of some salts or certain chemical agents.

30.

\(\log { \frac { { k }_{ 2 } }{ { k }_{ 1 } } } =\frac { { E }_{ a } }{ 2.303R } \left( \frac { { T }_{ 2 }-{ T }_{ 1 } }{ { T }_{ 1 }{ T }_{ 2 } } \right)\)
\(\frac { { k }_{ 2 } }{ { k }_{ 1 } } =2,\quad { T }_{ 1 }=298K,\quad { T }_{ 2 }=308K,\quad R=8.314\quad J{ K }^{ -1 }{ mol }^{ -1 }\)
\(\log { 2 } =\frac { { E }_{ a } }{ 2.303\times 8.314 } \times \left[ \frac { 308-298 }{ 308\times 298 } \right] \)
\(0.3010=\frac { { E }_{ a } }{ 2.303\times 8.314 } \times \frac { 10 }{ 298\times 308 } \)
\(\therefore \ { E }_{ a }=\frac { 0.3010\times 2.303\times 8.314\times 298\times 308 }{ 10 } \)
\(52.898 \ kJ\)

31.

We know from Kohlrausch law that
\(A_{m\left(\mathrm{CaCl}_{2}\right)}^{\mathrm{o}}=\lambda_{\mathrm{Ca}^{2+}}^{0}+2 \lambda_{\mathrm{Cl}^{-}}^{0}=119.0 \mathrm{~S} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}+2(76.3) \mathrm{S} \mathrm{cm}^{2} \mathrm{~mol}^{-1}\)
= (119.0 + 152.6) S cm² mol^–1
= 271.6 S cm² mol^–1
\(A_{n\left(\mathrm{MgSO}_{4}\right)}^{\mathrm{o}}=\lambda_{\mathrm{Mg}^{2+}}^{0}+\lambda_{\mathrm{so}_{4}^{2}}^{0}=106.0 \mathrm{~S} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}+160.0 \mathrm{~S} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}\)
= 266 S cm² mol^–1

Case Study Questions

3 x 4 = 12

32.

1.c) Fructose
2.(b) Epimers
3. (c) Glycosidic linkage

33.

1.(b) 0.174
2.(a) I
3.(c) 0.734
4.(a) ideal solution

34.

(i) (a)
(ii) (a)
(iii) (c) : Glucose contains four chiral centres.
(iv) (c) : Pentacetate of glucose does not react with hydroxylamine showing absence of free -CHO group. This cannot be explained by open structure of glucose.