Question

The vapour pressure of two liquids $P$ and $Q$ are $80$ and $60$ torr, respectively. The total vapour pressure of solution obtained by mixing $3$ moles of $P$ and $2$ moles of Q would be

• A. $72\text{torr}$
• B. $140\text{torr}$
• C. $68\text{torr}$
• D. $20\text{torr}$

Answer 1

The correct option is A $72\text{torr}$

By Raoult's Law,
$P_T=P_P^0{\chi }_P^0+P_Q^0{\chi }_Q^0$
Where, $P_T=\text{Total vapour pressure}$
Where $\text{Vapour pressure of P},P_P^0=80\text{torr}$,
$\text{Vapour pressure of Q},P_Q^0=60\text{torr}$
$\text{Mole fraction of P},{\chi }_P=\frac{3}{5}$
$\text{Mole fraction of Q},{\chi }_Q=\frac{2}{5}\therefore P_T=80\times \frac{3}{5}+60\times \frac{2}{5}=48+24=72\text{torr}$