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(a)<br />

(b)<br />

(c)<br />

(d)<br />

CH CH C CH<br />

CH CH C CH<br />

CH CH C CH<br />

C C—CH CH 2<br />

43. Determine the standard reduction potential for the<br />

half cell reaction, Cl 2 + 2e – 2Cl – .<br />

(Given : Pt 2+ + 2Cl – Pt + Cl 2 , E° cell = –0.15 V<br />

Pt 2+ + 2e – Pt, E° = 1.20 V)<br />

(a) 1.05 V<br />

(b) – 1.05 V<br />

(c) – 1.35 V (d) 1.35 V<br />

44. Among 2-chloropropanoic acid, 3-chloropropanoic<br />

acid, 2,2-dichloroacetic acid and propanoic acid,<br />

the K a values will be in the order,<br />

(a) 2,2-dichloroacetic acid > 2-chloropropanoic acid<br />

> 3-chloropropanoic acid > propanoic acid<br />

(b) 3-chloropropanoic acid > 2-chloropropanoic acid<br />

> 2,2-dichloroacetic acid > propanoic acid<br />

(c) 2,2-dichloroacetic acid > 3-chloropropanoic acid<br />

> 2-chloropropanoic acid > propanoic acid<br />

(d) 2,2-dichloroacetic acid > propanoic acid ><br />

2-chloropropanoic acid > 3-chloropropanoic acid<br />

45. Which one of the following is employed as a<br />

tranquilizer?<br />

(a) Naproxen (b) Tetracycline<br />

(c) Chlorpheniramine<br />

(d) Equanil<br />

SOLUTIONS<br />

1. (b): The de<strong>com</strong>position reaction of H 2 O 2 is<br />

2HO → 2HO+<br />

O<br />

2 2 2 2<br />

Thus, 2 mol (or 4 equivalents) of H 2 O 2 would give<br />

1 mol (22.4 L at STP) of O 2 . 1 L of 4 equivalents of<br />

H 2 O 2 has a volume strength of 22.4. Thus,<br />

22. 4×<br />

1.<br />

5<br />

1 L of 1.5 equivalents (1.5 N) of H 2 O 2 = = 8.4<br />

4<br />

2. (c) : Molarity of glucose in blood<br />

No.ofmoles of glucose<br />

=<br />

Volume of blood (inL)<br />

Wt.ofglucose<br />

=<br />

Mol. wt.ofglucose × Volume of blood (inL)<br />

[Mol. mass of glucose = 180 g/mol]<br />

09 . g<br />

=<br />

180 g/mol×<br />

1 L = 5 × 10–3 mol L –1 = 0.005 M<br />

3. (a)<br />

4. (a) : If ∆ o < P, then fourth electron will go to higher<br />

energy, e g orbital. Hence, the configuration be<strong>com</strong>es<br />

t 3 2g e 1 g.<br />

5. (b): For H-atom :<br />

1 1 1 <br />

= RH<br />

−<br />

2 2<br />

<br />

λ H <br />

n1<br />

n2<br />

<br />

...(i)<br />

For He + ion :<br />

1 <br />

2 1 1 <br />

= RH<br />

× Z −<br />

2 2<br />

...(ii)<br />

+ <br />

n<br />

1 n<br />

2 <br />

λ He<br />

1 1<br />

∴ λ + = λ<br />

He H<br />

× = 91. 2 × = 22.<br />

8 nm<br />

2 2<br />

Z 2<br />

( λ H = 91.2 nm)<br />

6. (b) : O 2 F 2 and H 2 O 2 , both have open book type<br />

structure.<br />

In O 2 F 2 , one O—O bond and two O—F bonds<br />

are lying in different planes, i.e., this molecule like<br />

H 2 O 2 has non-linear and non-planar structure.<br />

7. (d): Adding first two equations, we have,<br />

CaC2<br />

+ 2H2O+H2→ CaOH ( ) 2+<br />

C2H4<br />

64 g<br />

28 g<br />

i.e., 64 g of CaC 2 gives 28 g of C 2 H 4 .<br />

From 3 rd equation,<br />

nCH 2 4 → ( CH2CH2)<br />

n<br />

28n g of C 2 H 4 gives 28n g of polythene<br />

i.e., 28 g of C 2 H 4 gives 28 g of polythene.<br />

Hence, 64 g of CaC 2 will give 28 g of polythene or,<br />

64 kg of CaC 2 will give 28 kg of polythene.<br />

2+ 8. (a) : Mg NH3 HPO 2 −<br />

+ + 4 → MgNH ( 4)<br />

PO4<br />

White ppt.<br />

9. (c) : Greater the number of ions and greater the<br />

charge on each ion, greater will be the conductivity.<br />

The given <strong>com</strong>plexes ionise as,<br />

(P) Mg[Cr(NH 3 )(NO 2 ) 5 ]<br />

Mg 2+ + [Cr(NH 3 )(NO 2 ) 5 ] 2–<br />

No. of ions = 2<br />

14 CHEMISTRY TODAY | APRIL ‘17

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