The specific conductance of a 0.1 M KCl solution at 23$°\mathrm{C}$ is 0.012 ${\mathrm{\Omega }}^{-1}{\mathrm{cm}}^{-1}$.
The resistance of the cell containing the solution at the same temperature was found to be 55 $\mathrm{\Omega }$. The cell constant will be:

1. 0.142 cm^–1
2. 0.66 cm^–1
3. 0.918 cm^–1
4. 1.12 cm^–1

For the cell, Ti/Ti⁺(0.001M)||Cu²⁺(0.1M)|Cu, ${\mathrm{E}}_{\mathrm{cell}}^{\mathrm{o}}$ $$at

25 $°$C is 0.83 V. E_cell can be increased :

1. By increasing [Cu²⁺]

2. By increasing [Ti⁺]

3. By decreasing [Cu²⁺]

4. None of the above.

The unit of specific conductance is:

The metal that cannot be produced upon reduction of its oxide by aluminium is :

Adiponitrile is manufactured electrolytically from acrylonitrile. The reaction is as follows:
$C{H}_2=$ $CHCN$ $\to$ $CN$ $–$ $(C{H}_2$ ${)}_4$ $–$ $CN$

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

1. 30 kg                                               

2. 58 kg

3. 60 kg                                               

4. 80 kg

The useful work of the reaction \(\small{\mathrm{Ag}(\mathrm{s})+1 / 2 \mathrm{Cl}_2(\mathrm{~g}) \rightarrow \mathrm{AgCl}(\mathrm{s})}\) is: 

Given,
\(E_{Cl_{2}/Cl^{-}}^{o} \ = \ +1.36 \ V, \)
\(E_{AgCl/Ag, / Cl^{-}}^{o} \ = \ 0.22 \ V, \)
\( P_{Cl_{2}} \ = \ 1 \ atm\ and \)
\( T = 298 K\)

Consider the given cell:

Pt(s) | H₂(g, 1 bar) | H⁺ (0.030 M) || Br⁻ (0.010 M) | Br₂(l) | Pt(s)

If the concentration of Br⁻ ions is doubled and the concentration of H⁺ ions is reduced to half of its initial value,
how will the emf of the cell change?

1. Two times
2. Four times
3. Eight times
4. Remains the same

Aluminium metal can be produced by the electrolysis of molten aluminium oxide at about 1000 °C.
The cathode reaction is: \(Al^{3 +} + 3 e^{-} \rightarrow Al\)

Given that the atomic mass of aluminium is 27 amu and 1 Faraday = 96,500 C, calculate the quantity of electricity (in coulombs) required to produce 5.12 kg of aluminium by this method:

1. \(5 . 49 \times 10^{1 } C\) of electricity

2. \(5 . 49 \times 10^{4 } C\) of electricity

3. \(1 . 83 \times 10^{7 } C\) of electricity

4. \(5 . 49 \times 10^{7 } C\) of electricity

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  ${\Lambda }_{\mathrm{HA}}^{\infty }$ = 400 S ${\mathrm{cm}}^2$ ${\mathrm{mol}}^{-1}$]

Calculate the standard cell potential, E°cell, in volts and the standard Gibbs free-energy change, ΔG°, in kJ mol⁻¹, respectively, for a two-electron redox reaction at 298 K having an equilibrium constant of 3.8 × 10⁻³.