The vapour pressure of a saturated solution of sparingly soluble salt () was at . If the vapour pres — Solutions and Colligative Properties Chemistry Question
Question
The vapour pressure of a saturated solution of sparingly soluble salt ($\text{XCl}_3$) was $17.20 \text{ mm Hg}$ at $27^\circ\text{C}$. If the vapour pressure of pure water is $17.25 \text{ mm Hg}$ at $27^\circ\text{C}$, what is the solubility of the sparingly soluble salt $\text{XCl}_3$, in mole per litre?
💡 Solution & Explanation
Relative lowering of vapour pressure is $(P^\circ - P)/P = n_{total} / N_{solvent}$. $(17.25 - 17.20)/17.20 = 0.05/17.20 \approx 0.002907$. The salt $\text{XCl}_3$ dissociates into 4 ions ($\text{X}^{3+} + 3\text{Cl}^-$), so $i=4$. Assuming 1 L of water ($N_{solvent} \approx 55.55$ mol), we have $4n / 55.55 = 0.002907 \Rightarrow 4n \approx 0.1615 \Rightarrow n \approx 0.04037 \text{ mol/L} = 4.04 \times 10^{-2} \text{ M}$. Therefore, correct answer is A.