$\Lambda_{m(HAc)}^0$ is equal to . . . . . . .

The limiting molar conductivities of $X^{+}$ and $Y^{-2}$ ions are $45$ and $110 \ S \ cm^2 \ mol^{-1}$ respectively. The $\Lambda_{m}^{\infty}$ of $X_2Y$ is:

Resistance of a cell containing $0.02 \ M \ KCl$ solution is $164 \ \Omega$. If the cell is filled with $0.05 \ M \ AgNO_3$,the resistance becomes $75.8 \ \Omega$. Calculate the following: [Conductivity of $0.02 \ M \ KCl = 2.768 \times 10^{-3} \ \Omega^{-1} \ cm^{-1}$] $(i)$ Conductivity of $0.05 \ M \ AgNO_3$ (ii) Molar conductivity of $AgNO_3$ solution.

The unit $ohm^{-1}$ is used for

The limiting molar conductivities of the given electrolytes at $298 \ K$ follow the order: $\lambda^{\circ}_{(K^{+})} = 73.5, \lambda^{\circ}_{(Cl^{-})} = 76.3, \lambda^{\circ}_{(Ca^{2+})} = 119.0, \lambda^{\circ}_{(SO_{4}^{2-})} = 160.0 \ S \ cm^{2} \ mol^{-1}$.

The limiting molar conductivities of $NaI$,$NaNO_3$ and $AgNO_3$ are $12.7$,$12.0$ and $13.3 \, mS \, m^2 \, mol^{-1}$,respectively (all at $25^{\circ} C$). The limiting molar conductivity of $AgI$ at this temperature is $.... \, mS \, m^2 \, mol^{-1}$.