Question

The vapor pressure of benzene, c6h6, is 100.0 torr at 26.1 °c. Assuming raoult's law is obeyed, how many moles of a nonvolatile solute must be added to 100.0 ml of benzene to decrease its vapor pressure by 10.0% at 26.1 °c? The density of benzene is 0.8765 g>cm3.

Answer 1

According to Raoult's law:

$p_{solution} = X_{solvent}\times P_{solvent}$    -(1)

where $p_{solution}$ is observed vapor pressure of the solution, $X_{solvent}$ is mole fraction of solvent, and $P_{solvent}$ is vapor pressure of the pure solvent.

Vapor pressure of benzene = $100 torr at 26.1^{o} C$   (given)

Reduction in vapor pressure on addition of non-volatile solute = 10.0 %   (given)

So, vapor pressure of the component in the solution = $100 - 10 = 90 torr$

Substituting the values in formula (1):

$90 = mole fraction of benzene \times 100$

$mole fraction of benzene = \frac{90}{100} = 0.9$

Mole of benzene = $Volume\times \frac{density}{Molar Mass}$

Mole of benzene = $100 cm^{3}\times \frac{0.8765 g/cm^{3}}{78 g/mol} = 1.124 mol$

Since, $mole fraction of benzene =\frac{mole of benzene }{mole of benzene +mole of non volatile solute }$

So, $0.9 =\frac{1.124 }{1.124+mole of non volatile solute }$

$\frac{1.124}{0.9} =1.124+mole of non volatile solute$

$mole of non volatile solute = 1.249 - 1.124 = 0.125$

Hence, moles of a nonvolatile solute to be added is $0.125$.