The conductivity of 0.00241 M acetic acid is 7.896 × 10–5 S cm–1. If ${\mathrm{\Lambda }}_{\mathrm{m}}^{0}$ for acetic acid is 390.5 S cm2 mol–1, the dissociation constant will be
1. $$2.45 \times 10^{-5} \mathrm{~mol} \ \mathrm{~L}^{-1}$$
2. $$1.86 \times 10^{-5} \mathrm{~mol} \ \mathrm{L^{-1}}$$
3. $$3.72 \times 10^{-5}\mathrm{~mol} \mathrm{~L^{-1}}$$
4. $$2.12 \times 10^{-5}\mathrm{~mol} \mathrm{~L^{-1}}$$

Subtopic:  Conductance & Conductivity |
67%
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The resistance of a cell containing 0.001 M KCl solution at 298 K is 1500 . The conductivity is 0.146 × 10–3 S cm–1The cell constant would be-

$1.$ $0.12$ ${\mathrm{cm}}^{-1}$

$2.$ $0.56$ ${\mathrm{cm}}^{-1}$

$3.$ $0.22$ ${\mathrm{cm}}^{-1}$

$4.$ $1.36$ ${\mathrm{cm}}^{-1}$

Subtopic:   Kohlrausch Law & Cell Constant |
83%
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The amount of charge required for the reduction of 1 mol of ${\mathrm{MnO}}_{4}^{-}$ to ${\mathrm{Mn}}^{2+}$ is -

$1.$ $4.8$ $×$ ${10}^{5}$ $\mathrm{C}$
$2.$ $3.2$ $×$ ${10}^{6}$ $\mathrm{C}$
$3.$ $1.8$ $×$ ${10}^{5}$ $\mathrm{C}$
$4.$ $4.1$ $×$ ${10}^{4}$ $\mathrm{C}$

Subtopic:  Faraday’s Law of Electrolysis |
82%
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The number of Faradays required to produce 20.0 g of Ca from molten CaCl2 is-

1. 2F

2. 1F

3. 4F

4. 3F

Subtopic:  Faraday’s Law of Electrolysis |
75%
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Three electrolytic cells A, B, C containing solutions of ZnSO4, AgNO3, and CuSO4, respectively are connected in series.

A steady current of 1.5 amperes was passed through them until 1.45 g of silver was deposited at the cathode of cell B. The current flow time is-

1. 14 minutes

2. 25 minutes

3. 20 minutes

4. 11 minutes

Subtopic:  Faraday’s Law of Electrolysis |
60%
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Mg(s) | Mg2+(0.001M) || Cu2+(0.0001 M) | Cu(s)

The value of  for the above reaction is -

1. 3.46 V

2. 3.15 V

3. 2.67 V

4. 1.24 V

Subtopic:  Nernst Equation |
87%
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The correct increasing order of reducing power of the metals is:

 K+/K = –2.93 V Ag+/Ag = 0.80 V Hg2+/Hg = 0.79 V Mg2+/Mg = –2.37 V Cr3+/Cr = – 0.74 V

 1 Cr

Subtopic:  Electrode & Electrode Potential |
85%
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The correct statement about the given galvanic cell equation is -

Zn(s) + 2Ag+­­­(aq) → Zn2+(aq) + 2Ag(s)

 1 The current will flow from silver to zinc in the external circuit. 2 The current will flow from zinc to silver in the external circuit. 3 The current will flow from silver to zinc in the internal circuit. 4 The current will flow from zinc to silver in the internal circuit.

Subtopic:  Electrolytic & Electrochemical Cell |
60%
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2Cr(s) + 3Cd2+(aq)  2Cr3+(aq) + 3Cd

${{\mathrm{E}}^{⊝}}_{{\mathrm{Cr}}^{3+}/\mathrm{Cr}}=$ $-0.74$ $\mathrm{V}$${{\mathrm{E}}^{⊝}}_{{\mathrm{Cd}}^{2+}/\mathrm{Cd}}=-0.40$ $\mathrm{V}$

The value of $∆{\mathrm{G}}_{r}^{\mathrm{o}}$ in the above reaction will be-

 1 -196.83 kJ 2 196.83 kJ 3 186.83 kJ 4 -186.83 kJ

Subtopic:  Relation between Emf, G, Kc & pH |
74%
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The incorrect statement(s) among the below is/are:

 a. The unit of conductivity is S cm-2 b. Specific Conductivity of weak and strong electrolytes always decreases with a decrease in concentration. c. The unit of molar conductivity is S cm2 mol-1 d. Molar conductivity increases with an increase in concentration.

1. a

2. a and d

3. b and d

4. a and c

Subtopic:  Conductance & Conductivity |
57%
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