Question

$12g$ of non volatile solute having $(\text{Molar mass}=120g/mol)$ was dissolved in $9.9mol$ of solvent. If the vapour pressure of pure solvent is $P_A^{\circ },$ the vapour pressure of solution is:

• A. $0.99P_A^{\circ }$
• B. $0.01P_A^{\circ }$
• C. $0.12P_A^{\circ }$
• D. $0.88P_A^{\circ }$

Answer 1

The correct option is A $0.99P_A^{\circ }$

When a non volatile solute is added to a volatile solvent, vapour pressure is lowered.
By relative lowering of vapour pressure formula:

$\frac{P_A^{\circ }-P_S}{P_A^{\circ }}=x_{solute}\Rightarrow \frac{P_A^{\circ }-P_S}{P_A^{\circ }}=\frac{n_A}{n_A+n_B}$
$n_A\text{are the number of moles of solute}=\frac{12}{120}=0.1mol$
$n_B\text{are the number of moles of solvent}=9.9mol$
$\Rightarrow \frac{P_A^{\circ }-P_S}{P_A^{\circ }}=\frac{0.1}{0.1+9.9}\Rightarrow \frac{P_A^{\circ }-P_S}{P_A^{\circ }}=0.01\Rightarrow P_S=0.99P_A^{\circ }$