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. $\mathrm{x }- \mathrm{y }+ 2z$

2. $\mathrm{x }+ \mathrm{y }+ \mathrm{z }$

3. x - y + z

4. $\frac{(x-y)}{\mathrm{2 }}+ z$

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

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

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