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Chemistry NEET Questions — Free MCQ Practice

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Adiponitrile is manufactured electrolytically from acrylonitrile $C H_{2} = C H C N \to C N - (C H_{2})_{4} - C H$

How many kg of adiponitrile (molecular mass = 108) is produced in 9.65 hr using a current of 3750 A with 80% efficiency ?

A.

30 kg

B.

58 kg

C.

60 kg

D.

80 kg

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Explanation

(B) $2 C H_{2} = C H C N + 2 H^{+} + 2 e^{-} \to C N - (C H_{2})_{4} C N$

$m = \frac{(\frac{108}{2}) \times 10^{- 3} \times 3000 \times 9.65 \times 3600}{96500} = 58.32 k g$

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A certain metal salt solutions is electrolysed in series with a silver coulometer. The weights of silver and the metal deposited are 0.5094 g and 0.2653 g. Calculate the valency of the metal if its atomic weight is nearly that of silver.

A.

1

B.

2

C.

3

D.

4

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Explanation

(B) Evidently $\frac{0.5094}{0.2653} = \frac{E q u i v a l e n t w e i g h t o f A g}{E q u i v a l e n t w e i g h t o f m e t a l} \approx 0.2$

$∴ v a l e n c y r a t i o = \frac{V a l e n c y o f m e t a l}{V a l e n c y o f A g} = 2$

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The reduction potential of a half-cell consisting of a Pt electrode immersed in $1.5 M F e^{2 +}$ and $0.015 M F e^{3 +}$ solution at $25 ° C i s (E_{(A g^{+}, A g)}^{°} = + 0.770 v)$ .

A.

0.652 V

B.

0.88 V

C.

0.710 V

D.

0.850 V

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Explanation

(A) $E = 0.77 - \frac{0.059}{1} \log \frac{[F e^{2 +}]}{[F e^{3 +}]} = 0.77 - 0.059 \log \frac{1.5}{0.015} = 0.77 - 0.118 = 0.652 V$

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In a concentration cell, $Z n / Z n^{2 +} (1.0 M) | | Z n^{2 +} (0.15 M) / Z n$ as the cell discharges,

A.

reaction proceeds to the right

B.

the two solutions approach each other in concentration

C.

no reaction takes place

D.

water gets decomposed

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Explanation

(B). In the above representation of the cell, the reaction proceeds to the left; 1.0 M solution is used up.

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In a half-cell containing $[T l^{3 +}] = 0.1 M a n d [T l^{+}] = 0.01 M,$ the cell potential is – 1.2496 V for the reaction $T l^{+} \to T l^{3 +} + 2 e^{-}$ The standard reduction potential of the $T l^{+} / T l^{3 +}$ couple at $25 ° C$ is

A.

1.44 V

B.

0.61 V

C.

2.44 V

D.

1.22 V

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Explanation

(D) For the oxidation reaction,

$T l^{+} \to T l^{3 +} + 2 e^{-} E = E^{°} - \frac{R T}{n F} \log \frac{[o x i]}{[r e d]} - 1.2496 = E ° - \frac{0.059}{2} \log \frac{0.01}{0.01} ∴ E ° = 1.220 V$

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The standard reduction potentials for $M n^{3 +} / M n^{2 +}$ and $M n O_{2} / M n^{3 +}$ couples in acid medium at $25 ° C$ are 1.50 V and 1.00 V respectively. The following reaction:

$2 M n^{3 +} + 2 H_{2} O \to M n^{2 +} + M n O_{2} + 4 H^{+} i s$

A.

reversible

B.

non-spontaneous

C.

spontaneous

D.

irreversible

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Explanation

(C) $M n^{3 +} + e^{-} \to M n^{2 +}; E ° = 1.50 V M n^{3 +} + 2 H_{2} O \to M n O_{2} + 4 H^{+} + e^{-}; E ° = - 1.0 V$

Eº for the above reaction=1.50 V + (– 1.0)V = 0.50 V

Since Eº is positive, $∆$ Gº is negative and the reaction is spontaneous.

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If a 100 mL solution of 0.1M HBr is titrated using a very concentrated solution of NaOH, then the conductivity (specific conductance) of the solution at the equivalence point will be (assume volume change is negligible due to addition of NaOH). Calculate your answer in terms of $10^{- 1} S m^{- 1},$

$[G i v e n {\land^{°}}_{(N a^{+})} = 8 \times 10^{- 3} S m^{2} m o l^{- 1} {\land^{°}}_{(B r^{-})} = 4 \times 10^{- 3} S m^{2} m o l^{- 1}]$

A.

6

B.

12

C.

15

D.

24

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Explanation

(B) At the equivalence point the concentrations will be $[B r^{-}] = 100 m^{3}, [N a^{+}] = 100 m^{3}$

Therefore,

$K_{t o t a l} = k_{B r^{-}} + k_{N a^{+}} = 1.2 S m^{- 1} = 12 \times 10^{- 1} S m^{- 1}$

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The volume of gases liberated at STP when a charge of 2F is passed through aqueous solution of sodium phosphate, is :

A.

11.2L

B.

44.8L

C.

33.6L

D.

22.4L

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Explanation

(C) At anode $2 H_{2} O \to O_{2} + 4 H^{+} + 4 e^{-}$

At cathode $2 H_{2} O + 2 e^{-} \to 2 O H^{-} + H_{2}$

After passage of 2F, one mole of $H_{2}$ and 1/2 mole of $O_{2}$ would be produced. The total volume is 33.6L.

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It is desired to convert the energy derived from the combustion of propane into electrical energy, via a fuel cell. Given that the standard free energies of formation of $C_{3} H_{8} (g), H_{2} O (l) a n d C O_{2} (g)$ are -23.5, -237.2 and -394.4 $k J m o l^{- 1}$ respectively, calculate the standard EMF of the propane fuel cell.

A.

1.10 V

B.

2.10V

C.

3.12 V

D.

4.12 V

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Explanation

(A) For the combustion of propane

$C_{3} H_{8} + 5 O_{2} \to 3 C O_{2} (g) + 4 H_{2} O (l)$ the overall $∆ G °$ for the reaction

$= 3 x + 394.4 + 4 x - 237.2 = - [- 23.5 + 0] = - 1183.2 - 948.8 + 23.5 = - 2108.5 k J C_{3} H_{8} (g) + 6 H_{2} O (l) \to 3 C O_{2} (g) + 20 H = 20 e \frac{5 O_{2} (g) + 2 O H^{+} + 20 e \to 10 H 2 O (l)}{C_{3} H_{8} (g) + 5 O_{2} (g) \to 3 C O_{2} (g) + 4 H_{2} O (l)}$

$∴$ 20 electrons are involved in the cell reactions

$∆ G^{0} = - n F E^{0}; E^{0} = - \frac{∆ G^{0}}{n F} = \frac{+ 21085 \times 10^{3}}{20 \times 96500} = 1.10 V$

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Na-amalgam is prepared by electrolysis of NaCl solution using liquid Hg as cathode. How long should the current of 10 amp. is passed to produce 10% Na – Hg on a cathode of 10 gm Hg. (atomic mass of Na = 23).

A.

7.77 min.

B.

9.44 min.

C.

5.24 min.

D.

11.39 min.

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Explanation

(A) 90 gm Hg has 10 gm Na

$∴ 10 g m H g = \frac{10}{90} \times 10 = \frac{10}{9} g m N a ∴ w e i g h t o f N a = \frac{M}{n} \times \frac{i \times t}{96500} \frac{10}{9} = \frac{23}{1} \times \frac{10 \times t}{96500} [∴ N a^{+} + e \to N a] = \frac{10 \times 96500}{9 \times 10 + 23} = 7.77 m i n$

Chemistry MCQs for NEET — Practice Questions with Answers

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