Limiting molar conductivities, for the given solutions, are :

λm0(H2SO4)x cm2 mol-1

λm0(K2SO4)y cm2 mol-1

λm0(CH3COOK)z cm2 mol-1

From the data given above, it can be concluded that \(\lambda_m^0 \) in (\(S\ cm^2\ mol^{-1}\)) for CH3COOH will be :

1. \(\mathrm{x-y+2z}\)        2. \(\mathrm{x+y+z}\)          
3. \(\mathrm{x-y+z}\)        4. \(\mathrm{{(x-y) \over 2}+z}\)          

Subtopic:  Conductance & Conductivity |
 69%
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For the cell  PtsBr2lBr-0.010 MH+0.030 MH2g1 barPts,

If the concentration of Br- becomes 2 times and the concentration of H+ becomes half of the initial value, then emf of the cell

1.  Doubles

2.  Four times

3.  Eight times

4.  Remains the same

 

 

Subtopic:  Nernst Equation |
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The specific conductance (K) of 0.02 M aqueous acetic acid solution at 298 K is 1.65 × 10-4 S cm-1. The degree of dissociation of acetic acid is [Given: Equivalent conductance at infinite dilution of H+ = 349.1 S cm2 mol-1 and CH3COO- = 40.9 S cm2 mol-1)

1.  0.021

2.  0.21

3.  0.012

4.  0.12

Subtopic:  Conductance & Conductivity |
 64%
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The equilibrium constant of a 2 electron redox reaction at 298 K is 3.8 x 10-3. The cell potential Eo (in V) and the free energy change ∆Go (in kJ mol-1 ) for this equilibrium respectively, are -

1. -0.071, -13.8 2. -0.071, 13.8
3. 0.71, -13.8 4. 0.071, -13.8
Subtopic:  Relation between Emf, G, Kc & pH |
 62%
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The specific conductance of 0.01 M solution of a weak monobasic acid is 0.20 x 10-3 S cm-1. The dissociation constant of the acid is-

[Given  ΛHA = 400 S cm2 mol-1]

1. \(5 \times 10^{-2}\) 2. \(2.5 \times 10^{-5}\)
3. \(5 \times 10^{-4}\) 4. \(2.2 \times 10^{-11}\)
Subtopic:  Conductance & Conductivity |
 56%
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Equivalent conductance of saturated BaSO4 solution is 400 ohm-1 cm2 equivalent-1 and it's specific conductance is 8 × 10-5 ohm-1 cm-1; hence solubility product Ksp of BaSO4 is : 

1.  4 × 10-8 M2

2.  1 × 10-8 M2

3.  2 × 10-4 M2

4.  1 × 10-4 M2

Subtopic:  Conductance & Conductivity |
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Aluminium oxide may be electrolysed at 1000 C to furnish aluminium metal (Atomic mass = 27 amu; 1 Faraday = 96,500 Coulombs). The cathode reaction is: Al3++3e-Al

To prepare 5.12 kg of aluminium metal by this method, would require : 

1. 5.49×10C of electricity 

2. 1.83×10C of electricity

3. 5.49×10C of electricity 

4. 5.49×10C of electricity

Subtopic:  Faraday’s Law of Electrolysis |
 61%
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The potential of a hydrogen electrode having a pH = 10 is : 

1. 0.59 V

2. –0.59 V

3.  0.0 V

4. –5.9 V

Subtopic:  Nernst Equation |
 68%
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Calculate the Emf of the given cell: 

Zn(s) | Zn+2 (0.1M) || Sn+2 (0.001M) | Sn(s)

(Given EZn+2/Zno=-0.76 V, ESn2+/Sno=-0.14 V)

1. 0.62 V

2. 0.56 V

3. 1.12 V

4. 0.31 V

Subtopic:  Electrode & Electrode Potential |
 56%
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For a reaction A(s) + 2B+  A2+ + 2B(s) ;  KC has been found to be 1012. The Ecell° is : 

1. 0.35 V 2. 0.71 V
3. 0.01 V 4. 1.36 V
Subtopic:  Relation between Emf, G, Kc & pH |
 77%
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