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 $8 \times {10}^{-5} {\mathrm{ohm}}^{-1} {\mathrm{cm}}^{-1}$; hence solubility product ${\mathrm{K}}_{\mathrm{sp}} \mathrm{of} {\mathrm{BaSO}}_4$ is : 

1.  $4 \times {10}^{-8} {\mathrm{M}}^2$

2.  $1 \times {10}^{-8} {\mathrm{M}}^2$

3.  $2 \times {10}^{-4} {\mathrm{M}}^2$

4.  $1 \times {10}^{-4} {\mathrm{M}}^2$

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: ${\mathrm{Al}}^{3+}+3{\mathrm{e}}^{-}\to \mathrm{Al}$

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

1. $5.49\times {10}^{\mathrm{7 }}C$ of electricity 

2. $1.83\times {10}^{\mathrm{7 }}C$ of electricity

3. $5.49\times {10}^{\mathrm{4 }}C$ of electricity 

4. $5.49\times {10}^{\mathrm{9 }}C$ 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 ${\mathrm{cm}}^2$ ${\mathrm{mol}}^{-1}$]

The equilibrium constant of a 2 electron redox reaction at 298 K is 3.8 x ${10}^{-3}$. The cell potential E^o (in V) and the free energy change ∆G^o (in kJ mol^-1 ) for this equilibrium respectively, are -

The specific conductance (K) of 0.02 M aqueous acetic acid solution at 298 K is $1.65 \times {10}^{-4}$ 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 \(Pt(s)|| Br^-(aq)(0.010M)| Br_2(l)|| H^+(aq) (0.0030M) |H_2(g) (1 bar) | Pt(s)\)

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

1.  Doubles

2.  Four times

3.  Eight times

4.  Remains the same

Limiting molar conductivities, for the given solutions, are :

${\lambda }_m^0\left(H_{2}S{O}_4\right)= x$ $\mathrm{S }c{m}^2$ $mo{l}^{-1}$

${\lambda }_m^0\left(K_{2}S{O}_4\right)= y$ $\mathrm{S }c{m}^2$ $mo{l}^{-1}$

${\lambda }_m^0\left(C{H}_{3}COOK\right)= z$ $\mathrm{S }c{m}^2$ $mo{l}^{-1}$

From the data given above, it can be concluded that \(\lambda_m^0 \) in (\(S\ cm^2\ mol^{-1}\)) for CH₃COOH will be :
1. \(\mathrm{x-y+2z}\)       
2. \(\mathrm{x+y+z}\)          
3. \(\mathrm{x-y+z}\)       
4. \(\mathrm{{(x-y) \over 2}+z}\)          

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