\(\land^o_m\) 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:

Limiting molar conductivities, for the given solutions, are :

${\lambda }_m^0\left(H_{2}S{O}_4\right)=\hspace{0.17em}x$ $\mathrm{S }c{m}^2$ $mo{l}^{-1}$

${\lambda }_m^0\left(K_{2}S{O}_4\right)=\hspace{0.17em}y$ $\mathrm{S }c{m}^2$ $mo{l}^{-1}$

${\lambda }_m^0\left(C{H}_{3}COOK\right)=\hspace{0.17em}z$ $\mathrm{S }c{m}^2$ $mo{l}^{-1}$

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