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 ${\mathrm{cm}}^2$ ${\mathrm{mol}}^{-1}$]

1.	\(5 \times 10^{-2}\)	2.	\(2.5 \times 10^{-5}\)
3.	\(5 \times 10^{-4}\)	4.	\(2.2 \times 10^{-11}\)

Equivalent conductance of saturated \(\mathrm {BaSO}_4\) solution is \(400 \mathrm { ~ohm}^{-1}\) \(\mathrm {cm}^2\)  \(\mathrm { ~equivalent}^{-1}\) and it's specific conductance is \(8 \times 10^{-5} \text { ohm}^{-1} \text {cm}^{-1}\) ; hence solubility product \(K_{sp}\) of \(\mathrm {BaSO}_4\) is :

1. \(4 \times 10^{-8} \text {M}^2\)
2. \(1 \times 10^{-8} \text {M}^2\)
3. \(2 \times 10^{-4} \text {M}^2\)
4. \(1 \times 10^{-4} \text {M}^2\)

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

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.

The specific conductance (K) of 0.02 M aqueous acetic acid solution at 298 K is $1.65<mspace\; width="1em"></mspace>\times <mspace\; width="1em"></mspace>{10}^{-4}$ S ${\mathrm{cm}}^{-1}$. The degree of dissociation of acetic acid is [Given: Equivalent conductance at infinite dilution of ${\mathrm{H}}^{+}$ = 349.1 S ${\mathrm{cm}}^2$ ${\mathrm{mol}}^{-1}$ and ${\mathrm{CH}}_3{\mathrm{COO}}^{-}$ = 40.9 S ${\mathrm{cm}}^2$ ${\mathrm{mol}}^{-1}$)

1.  0.021

2.  0.21

3.  0.012

4.  0.12

$$
For the Cell \(\mathrm{Pt(s)|| Br^-(aq)(0.010M)| Br_2(l)|| H^+(aq) (0.030M) |H_2(g) (1 bar) | Pt(s)}\)

If the concentration of ${\mathrm{Br}}^{-}$ becomes 2 times and the concentration of ${\mathrm{H}}^{+}$ becomes half of the initial value, then emf of the cell is:

1.  Doubles

2.  Four times

3.  Eight times

4.  Remains the same

Limiting molar conductivities, for the given solutions, are:

\(\lambda_{m}^{0} \left(\right. H_{2} S O_{4} \left.\right) = x\) \(S  c m^{2}\) \(m o l^{- 1}\)

\(\lambda_{m}^{0} \left(\right. K_{2} S O_{4} \left.\right) = y\) \(S  c m^{2}\) \(m o l^{- 1}\)

\(\lambda_{m}^{0} \left(\right. C H_{3} C O O K \left.\right) = z\) \(S  c m^{2}\) \(m o l^{- 1}\)

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

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

Calculate the emf of the given cell: 

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

(Given ${\mathrm{E}}_{{\mathrm{Zn}}^{+2}/\mathrm{Zn}}^{\mathrm{o}}=-0.76$ $\mathrm{V},$ ${\mathrm{E}}_{{\mathrm{Sn}}^{2+}/\mathrm{Sn}}^{\mathrm{o}}=-0.14$ $\mathrm{V)}$

1. 0.62 V

2. 0.56 V

3. 1.12 V

4. 0.31 V

The electrode potential of Cu electrode dipped in 0.025 M CuSO₄ solution at 298 K is:

(standard reduction potential of Cu = 0.34 V)

1. 0.047 V

2. 0.293 V

3. 0.35 V

4. 0.387 V