Question

A saturated solution of sparingly soluble salt $M{Cl}_2$ has a vapour pressure of $31.78mm$ of $Hg$ at ${30}^{o}C$. Pure water exerts a pressure of $31.82mm$ of $Hg$ at ${30}^{o}C$. The $K_{sp}$ of $M{Cl}_2$ is
Assume molarity equal to molality and $M{Cl}_2$ is $100$% dissociated in solution.

• A. $1.3\times {10}^{-4}M$
• B. $2.6\times {10}^{-4}M$
• C. $1.7\times {10}^{-4}M$
• D. $5.8\times {10}^{-7}M$

Answer 1

The correct option is C $1.7\times {10}^{-4}M$

Assume 1 L of solution. It contains 55.56 moles of water
$\frac{P^0-P}{P^0}=\frac{n_{MC{l}_2}}{n_{water}}$

$\frac{31.82-31.78}{31.82}=\frac{n_{MC{l}_2}}{55.56}$

$n_{MC{l}_2}=6.98\times {10}^{-2}$

Since these moles are present in 1 L, they account for the molarity of solution.

But since $MC{l}_2$ is 100% ionized in solution, it gives 3 ions per molecule.
Hence, the solubility will one-third of $6.98\times {10}^{-2}$ which comes out to be $0.0349M$.

The expression for the solubility product is
$K_{sp}=\left[M^{2+}\right][C{l}^{-}{]}^2=(S\left)\right(2S{)}^2=4S^3$

$K_{sp}=4\times (0.349{)}^3=1.7\times {10}^{-4}$