Chemistry - Liquid Solution Question with Solution | TestHub

Given: $n_A=4$, $n_B=6$

Mole fraction of A:

$X_A=\frac{4}{4+6}=0.4$, $X_B=\frac{6}{10}=0.6$

According to Raoult's law:

$P_{total}=X_A{P}_A^0+X_B{P}_B^0$

At the normal boiling point, the vapor pressure of the solution equals the atmospheric pressure, which is $760mmHg$.

$760=0.4(500)+0.6P_B^0$

$760=200+0.6P_B^0$

$\Rightarrow P_B^0=\frac{560}{0.6}=933.3mmHg$

Therefore, the vapor pressure of pure B is $933mmHg$.