Question

A saturated solution of $XC{l}_3$ has a vapour pressure $17.20$ mm Hg at ${20}^{\circ }C$, while pure water vapour pressure is $17.25$ mm Hg. Solubility product $\left(K_{sp}\right)$ of $XC{l}_3$ at ${20}^{\circ }C$ is

• A. $1.8\times {10}^{-2}$
• B. ${10}^{-5}$
• C. $2.56\times {10}^{-6}$
• D. $7\times {10}^{-5}$

Answer 1

The correct option is A $1.8\times {10}^{-2}$

The relationship between the vapour pressure of solution and vapour pressure of pure solvent is $p=p^{0}x$.
Thus, $17.20=17.25x$
The mole fraction of water is $x=0.9971$
Hence, the mole fraction of $XC{l}_3$ is $1-x=1-0.9971=0.002899$
But $0.002899=\frac{n}{55.55+n}=\frac{n}{55.55}$ for 1 L of water $⟹n=0.161$ moles
The concentration of $XC{l}_3$ is $0.161$ M.
The expression for the solubility product of $XC{l}_3$ is $K_{SP}=\left[X^{3+}\right][C{l}^{-}{]}^3=(S\left)\right(3S{)}^3=27S^4$
Substitute values in the above expression, we get
$K_{SP}=27(0.161{)}^4=0.0181$