Question

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

• A. $9.80\times {10}^{-2}$
• B. $9.45\times {10}^{-6}$
• C. $2.56\times {10}^{-6}$
• D. $6.91\times {10}^{-5}$

Answer 1

The correct option is D $6.91\times {10}^{-5}$

We already know that,
$m=\frac{x_{solute}}{x_{solvent}}\times \frac{1000}{M_{Solvent}}=\frac{P°-P}{P}\times \frac{1000}{M_{Solvent}}$
As the solution is saturated. So, we can assume complete dissociation.$\mathrm{mi}=\frac{P°-P}{P}\times \frac{1000}{M_{\mathrm{solvent}}}$
$4.m=\frac{(17.25-17.20)\times 1000}{17.2\times 18}$
m= 0.040
for dilute solution molality is aproximately equal to molarity or S.
$\underset{1-S}{{\mathrm{XCl}}_3}\rightleftharpoons \underset{S}{X^{+}\left(\mathrm{aq}\right)+}3\underset{3S}{{\mathrm{Cl}}^{-}\left(\mathrm{aq}\right)}$
$K_{sp}$ = 27 $S^4$ = $6.91\times {10}^{-5}$