\(\land^o_m\) for NaCl, HCl and \(\mathrm{CH_3COONa }\) 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:
1. 0.01233
2. 1.00
3. 0.1233
4. 1.233

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

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