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, if the solution behaves ideally:

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

Answer 1

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

$\text{Given,}\text{Vapour pressure of pure liquid P}{P}_P^o=80\text{torr;}\text{Vapour pressure of pure liquid Q}{P}_Q^o=60\text{torr}$
$\text{Moles of P}\left(n_P\right)=3\text{mol ;}\text{Moles of Q}\left(n_Q\right)=2\text{mol}$
$\therefore \text{Total moles}\left(n_T\right)=3+2=5\text{mol}$
$\text{mole fraction of P}\left({\chi }_P\right)=\frac{n_P}{n_T}=\frac{3}{5}=0.6;$
$\therefore \text{Mole fraction of Q}\left({\chi }_Q\right)=1-{\chi }_P=1-0.6=0.4$
$\text{According to Raoult's law}$
$P_T=P_P^o{\chi }_P+P_Q^o{\chi }_Q$
$=80\times 0.6+60\times 0.4$
$48+24=72\text{torr}$