Question

The solubility product of $AgCl$ is ${10}^{-10}{M}^2$. The minimum volume (in $m^3$) of water required to dissolve 14.35 mg of $AgCl$ is approximately:

• A. 0.01
• B. 0.1
• C. 100
• D. 10

Answer 1

The correct option is D 10

$AgCl\rightleftharpoons {Ag}^{+}+{Cl}^{-}$

$S$ $S$

$\Rightarrow$ $K_{sp}=S^2$

$\Rightarrow$ ${10}^{-10}=S^2;S={10}^{-5}M$

$\Rightarrow$ Molecular weight of $AgCl=143.5g$

$\Rightarrow {10}^{-5}=\frac{14.35\times {10}^{-3}}{143.5}\times \frac{1000}{V\left(ml\right)}$

${10}^{-1}=\frac{1000}{V\left(ml\right)}$

$V_{ml}=10000ml=10$ litres

Ans :- D