Question

The solubility product of a sparingly soluble salt $A_2{X}_3$ is $1.1\times 10^{-23}$. If specific conductance of the solution is $3\times 10^{-5}S{m}^{-1}$, the limiting molar conductivity of the solution is $x\times 10^{-3}S{m}^{2}mo{l}^{-1}$. The value of $x$ is _____

Answer 1

Answer: 3

$s=10^{-5}M=10^{-5}\frac{mol}{L}=0.01\frac{mol}{m^3}$
Now ${\wedge }_m={\wedge }_m^{\infty }=\frac{k}{m}=\frac{k}{s}$
$\Rightarrow {\wedge }_m^{\infty }=\frac{3\times 10^{-5}}{0.01}=3\times 10^{-3}S-m^2/mol$