What is the conductivity of $0.05 \ M \ BaCl_2$ solution if its molar conductivity is $220 \ \Omega^{-1} \ cm^2 \ mol^{-1}$?

At $298 \ K$,the equivalent conductance of $0.1 \ M$ acetic acid is $5.20 \ S \ cm^{2} \ eq.^{-1}$. Calculate the degree of dissociation of acetic acid at this concentration in $\%$. Given: $\lambda^{\infty} (H^{+}) = 349.8 \ S \ cm^{2} \ mol^{-1}$ and $\lambda^{\infty} (CH_{3}COO^{-}) = 40.9 \ S \ cm^{2} \ mol^{-1}$.

The molar conductivities of $KCl$,$NaCl$,and $KNO_3$ are $152$,$128$,and $111 \ S \ cm^2 \ mol^{-1}$ respectively. The molar conductivity of $NaNO_3$ is ............. $S \ cm^2 \ mol^{-1}$.

The conductivity of sodium chloride at $298 \ K$ has been determined at different concentrations and the results are given below:

Concentration $/ M$	$0.001$	$0.010$	$0.020$	$0.050$	$0.100$
${10^2} \times \kappa / S \, m^{-1}$	$1.237$	$11.85$	$23.15$	$55.53$	$106.74$

Calculate ${\Lambda _m}$ for all concentrations and draw a plot between ${\Lambda _m}$ and $c^{1/2}$. Find the value of $\Lambda _m^o$.

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 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: