The resistance of a $0.1 \, N$ solution of a salt is $2.5 \times 10^{3} \, \Omega$. If the cell constant is $1.15 \, cm^{-1}$,what will be the equivalent conductance of the solution in $\Omega^{-1} \, cm^{2} \, eq^{-1}$ (in $.6$)?

Molar conductivity of a weak acid $HQ$ of concentration $0.18 \ M$ was found to be $1/30$ of the molar conductivity of another weak acid $HZ$ with concentration of $0.02 \ M$. If $\lambda_{Q^{-}}^0 = \lambda_{Z^{-}}^0$, then the difference of the $pK_a$ values of the two weak acids $(pK_a(HQ) - pK_a(HZ))$ is . . . . . . (Nearest integer).
[Given: degree of dissociation $(\alpha)$ $\ll 1$ for both weak acids, $\lambda^0$: limiting molar conductivity of ions]

Molar conductivities $(\Lambda ^o_m)$ at infinite dilution of $NaCl$,$HCl$ and $CH_3COONa$ are $126.4$,$425.9$ and $91.0 \ S \ cm^2 \ mol^{-1}$ respectively. $(\Lambda ^o_m)$ for $CH_3COOH$ will be .......... $S \ cm^2 \ mol^{-1}$.

The conductivity of a $0.05 \, M$ solution of a weak monobasic acid is $10^{-3} \, S \, cm^{-1}$. If $\lambda_{m}^{\infty}$ for the weak acid is $500 \, S \, cm^{2} \, mol^{-1}$,calculate the $K_{a}$ of the weak monobasic acid.

The molar conductivities $\Lambda_{NaOH}^o$ and $\Lambda_{HCl}^o$ at infinite dilution in water at $25 ^\circ C$ are $91.0$ and $426.2 \ S \ cm^2/mol$ respectively. To calculate $\Lambda_{CH_3COOH}^o$,the additional value required is:

Resistance of a conductivity cell (cell constant $129 \; m^{-1}$) filled with $74.5 \; ppm$ solution of $KCl$ is $100 \; \Omega$ (labelled as solution $1$). When the same cell is filled with $KCl$ solution of $149 \; ppm$,the resistance is $50 \; \Omega$ (labelled as solution $2$). The ratio of molar conductivity of solution $1$ and solution $2$ is i.e.,$\frac{\wedge_{1}}{\wedge_{2}} = x \times 10^{-3}$. The value of $x$ is (Nearest integer). Given,molar mass of $KCl$ is $74.5 \; g \; mol^{-1}$.