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. $\frac{(x-y)}{\mathrm{2 }}+ z$

\(\land^o_m\) 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-
1. 0.01233
2. 1.00 
3. 0.1233 
4. 1.233