correct answer is 25

# Explanation

$Mn{O}_4^{-}+8H^{+}+5e^{-}\to M{n}^{+2}+4H_{2}O,E^o=1.51V$

1 mole of MnO₄^- required 5 moles of electrons or 5 F electricity.

$\therefore$ 5 moles of MnO₄^- required 25 F electricity.

Explanation

Correct answer is C

Millimoles of Au⁺ = 0.1 × 250 = 25

Mole of Au⁺ = $\frac{25}{1000}$ = $\frac{1}{40}$

Charge passed = I × t = 1 × 15 × 60 = 900 C

moles of e^– passed = $\frac{900}{96500}$ = $\frac{9}{965}$

Only gold will be deposited as quantity of charge passed is less than the amount of Au⁺ present.

correct answer is 11

# Explanation

For synthesis of 1 mole of ClO₃^- , 6F of charge is required.

$\therefore$ To synthesise $\frac{10}{122}$ moles of KClO₃,

Charge required = $\frac{10}{122}\times 6$ F

$\frac{10}{122}\times 6=\frac{2\times t(hr)\times 3600}{96500}\times \frac{60}{100}$

$\Rightarrow$ t(hr) = $\frac{965\times 100}{122\times 2\times 36}$ = 10.98 hr $\simeq$ 11 Hr

correct answer is 60

# Explanation

Cr₂O₇^2- + 14H⁺ + 6e^– $\to$ 2Cr³⁺ + 7H₂O

I = 2 A, t = 8 min

From 1^st law of faraday,

w_Cr⁺³ = z $\times$ i $\times$ t

$\Rightarrow$ w_Cr⁺³ = $\frac{52}{96000\times 3}\times 2\times 8\times 60$

= $\frac{52}{300}$

Mass of Cr³⁺ ions actually obtained = 0.104 gm

% efficiency =

$\times$ 100

= $\frac{0.104}{\frac{52}{300}}$ $\times$ 100

= 60 %

correct answer is 5.66 To 5.68

# Explanation

Cathode : Ag⁺(aq) + e^- $\to$ Ag(s)

Moles of Ag deposited = $\frac{108}{108}$ = 1 mole

Anode : 2H₂O $\to$ O₂ + 4H⁺ + 4e^-

Here we have to find volume of O₂ evolved.

Equivalance of Ag = Equivalance of O₂

$\Rightarrow$ 1 $\times$ 1 = n_O2 $\times$ 4

$\Rightarrow$ n_O2 = $\frac{1}{4}$ mol

$\therefore$ Volume of O₂ evolved

= $\frac{1}{4}$ $\times$ 22.4

= 5.6 lit

Explanation

Correct answer is D

Cathode reaction :

Ni⁺² + 2e^- $\to$ Ni(s)

$\therefore$ From 2 mole of electrons 1 mole of Ni is deposited at the cathode.

So from 0.1 F or 0.1 mole of electrons $\frac{1}{2}\times 0.1$ = 0.05 mole of Ni is deposited at the cathode.

Explanation

Correct answer is C

Pb(s) + SO${}_4^{-2}$  $\to$  PbSO₄ + 2e^$-$

$\frac{n_{PbS{O}_4}}{1}=\frac{n_{e-}}{2}$

$\Rightarrow$   $\frac{n_{PbS{O}_4}}{1}=\frac{0.05}{2}$

$\therefore$  Weight of PbSO₄ $=$ $\frac{0.05}{2}\times 303=7.6g$

Explanation

Correct answer is A

Moler mass of p$-$aminophenol

=   6 $\times$ 12 + 7 + 14 + 16

= 109 gmol^$-$1

Eq. weight (E) = $\frac{W}{Q}$ $\times$ 96500

$\Rightarrow$$$ W = $\frac{EQ}{96500}$

$\Rightarrow$$$ W = $\frac{EIt}{96500}$

$\Rightarrow$$$ W = $\frac{109}{4}\times \frac{9.65\times 3600}{96500}$

$\Rightarrow$$$ W = 9.81 gm

Explanation

Correct answer is D

Required reaction :

B₂H₆ + 3O₂ $\to$ B₂ O₃ + 3 H₂ O

Here molar mass of B₂H₆ =10.8 $\times$ 2 + 6 = 27.6 gm

Given weight of B₂H₆ = 27.66 g

$\therefore$No of moles of B₂H₆ = $\frac{27.6}{27.66}\simeq 1$ mole.

For combustion of 1 mole B₂H₆ 3 moles O₂ required.

This 3 mole of O₂ is obtained by electrolysis of H₂O.

2H₂O($l$) $\to$ O₂ (g) + 4 H⁺ (aq) + 4 e^$-$

From Faradays law of electrolysis,

moles $\times$ n_f = $\frac{It}{96500}$

Here moles of O₂ = 3.

N_f of O₂ = 4 (in H₂ change of O = $-$2

and in O₂ change of 0 = O.

So change in charge = 2 .

for two atoms of O₂ change in charge = 2 $\times$ 2 = 4)

$\therefore$ 3 $\times$ 4 = $\frac{100\times t}{96500}$

$\Rightarrow$ t = 12 $\times$ 965 sec.

$\Rightarrow$ t = $\frac{12\times 965}{60\times 60}$ hr

= 3.2 hr

Explanation

Correct answer is A

Reaction at cathode :

2H⁺ + 2e^$-$ $\to$ H₂

We know,

$\omega$ = zIt = $\frac{EIt}{96500}$
<
no. of moles of H₂ = $\frac{112}{22400}$

$\therefore$ mass (w) of H₂ = $\frac{112}{22400}$ $\times$ 2

$\therefore$ $\frac{112}{22400}$ $\times$ 2 = $\frac{1\times I\times 965}{96500}$

$\Rightarrow$$$ $I$ = 1 A