Question

The vapour pressure of a saturated solution of sparingly soluble salt $\left(XC{l}_3\right)$ was 17.20 mm Hg at ${27}^{0}C$.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/Litre?

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

Answer 1

The correct option is A $4.04\times {10}^{-2}$

Given

Vapour pressure of solution of salt 17.20mmHg

Vapour Pressure of pure water 17.25mmHg

Solute is $XC{l}_3$

Solvent is water

Solution

$\frac{P_0-P_s}{P_s}=\frac{n}{N}$

n is moles of solute

N is moles of solvent

1L of water have 55.56 moles

N=55.56

Putting values

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

n=0.161

Solute will dissociate into 4 ions

Solubility of Solute is 0.161/4

$4.04\ast {10}^{-2}$

The correct option is A