Chemistry - Liquid Solution Question with Solution | TestHub

Given:
Mass of solute, $w_2=2$ g
Molar mass of solute, $M_2=64$ g mol${}^{-1}$
Mass of solvent (benzene), $w_1=50$ g
Molar mass of solvent, $M_1=78$ g mol${}^{-1}$
Vapour pressure of pure benzene, $P_1^0=640$ mmHg

Number of moles of solute, $n_2=\frac{w_2}{M_2}=\frac{2}{64}=\frac{1}{32}$ mol

Number of moles of solvent, $n_1=\frac{w_1}{M_1}=\frac{50}{78}=\frac{25}{39}$ mol

Mole fraction of solvent, $x_1=\frac{n_1}{n_1+n_2}$

$x_1=\frac{\frac{25}{39}}{\frac{25}{39}+\frac{1}{32}}=\frac{25\times 32}{25\times 32+39}=\frac{800}{800+39}=\frac{800}{839}$

According to Raoult's law, the vapour pressure of the solution is:
$P_1=P_1^0{x}_1$

$P_1=640\times \frac{800}{839}=\frac{512000}{839}\approx 610.25$ mmHg

Answer: 610.25 mm Hg