Question

The vapour pressure of a saturated solution of sparingly soluble salt $\left(XC{l}_3\right)$ was 17.20 mm Hg at ${27}^{\circ }C$. If the vapour pressure of pure $H_{2}O$ is 17.25 mm Hg at 300 K, what is the solubility of sparingly soluble salt $XC{l}_3$ in $mole/L$?

• A. $4.02\times {10}^{-2}$
• B. $8.08\times {10}^{-2}$
• C. $2.02\times {10}^{-2}$
• D. $4.04\times {10}^{-3}$

Answer 1

Answer: (A)

$4.02\times {10}^{-2}$

The exact mathematical expression of Raoult's law is $\frac{P^0-P_s}{P_s}=\frac{n}{N}$

Here, $P^0$ represents the vapour pressure of pure solvent, P represents the vapour pressure of the solution, n represents the number of moles of solute and N represents the number of moles of the solvent.

1 L of water contains 55.56 moles

Substitute values in the above expression.

$\frac{17.25-17.20}{17.25}=\frac{n}{55.56}$

Thus $n=0.161$

One molecule of $XC{l}_3$ dissociates to give 4 particles.

Hence, the solubility of $XC{l}_3$ is $\frac{0.161}{4}=4.02\times {10}^{-2}$

Hence, the correct option is $A$