\(\land^o_m\) for NaCl, HCl and \(CH_3COONa \) are 126.4, 425.9, and 91.05 S cm² mol^-1 respectively. If the conductivity of 0.001028 mol L^-1 acetic acid solution is \(4.95 \times 10^{-5} S ~cm^{-1} \), the degree of dissociation of the acetic acid solution is-

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 :

The molar conductivity of a 0.5 mol/dm³ solution of AgNO₃ with electrolytic conductivity of 5.76 x 10^-3 S cm^-1 at 298 K is:

1. 11.5 S cm²/mol
2. 21.5 S cm²/mol
3. 31.5 S cm²/mol
4. 41.5 S cm²/mol

The correct expression that  represents the equivalent conductance at infinite dilution of $A{l}_2{\left(S{O}_4\right)}_3$ is:

(Given that ${\wedge }_{A{l}^{3+}}^{°}$ $and$ ${\wedge }_{S{O}_4^{2-}}^{°}$ are the equivalent conductances at infinite dilution of the respective ions)

1. ${\wedge }_{A{l}^{3+}}^{°}$ $+$ ${\wedge }_{S{O}_4^{2-}}^{°}$

2. $\left({\wedge }_{A{l}^{3+}}^{°}\right)$ $+$ ${\wedge }_{S{O}_4^{2-}}^{°}\times 6$

3. $\frac{1}{3}{\wedge }_{A{l}^{3+}}^{°}$ $+\frac{1}{2}$ ${\wedge }_{S{O}_4^{2-}}^{°}$

4. $2{\wedge }_{A{l}^{3+}}^{°}$ $+3$ ${\wedge }_{S{O}_4^{2-}}^{°}$