Question

The specific conductance of a saturated solution of silver chloride is $K{\Omega }^{-1}c{m}^{-1}$. The limiting ionic conductances of $Ag$${}^{+}$and $Cl$${}^{-}$ ions are $x$ and $y$, respectively. The solubility of silver chloride in g litre${}^{-1}$ is :

[Molar mass of $AgCl=143.5$]

• A. $\frac{k\times 1000}{x-y}$
• B. $\frac{k}{x+y}\times 143.5$
• C. $\frac{k\times 1000\times 143.5}{x+y}$
• D. $\frac{x+y}{k}\times \frac{1000}{143.5}$

Answer 1

Answer: (C)

$\frac{k\times 1000\times 143.5}{x+y}$

The molar conductance of AgCl solution is ${\Lambda }_{AgCl}^m={\lambda }_{Ag+}^m+{\lambda }_{Cl-}=x+y$.

The relationship between molar conductance and specific conductance is ${\Lambda }_{AgCl}^m=\frac{1000\times \kappa }{C}$

Hence,

$x+y=\frac{1000\times \kappa }{C}C\frac{mol}{L}=\frac{1000\times \kappa }{x+y}$.

Now convert the concentration from $\frac{mol}{L}$ to $\frac{g}{L}$.
$C\frac{mol}{L}\times 143.5\frac{g}{mol}=143.5\times C\frac{g}{L}=143.5\times \frac{1000\times \kappa }{x+y}\frac{g}{L}.$

Hence, option C is correct.