Question

At a certain temperature, the vapor pressure of pure benzene_{$\left(C_6{H}_6\right)$} is 0.930 atm. A solution was prepared by dissolving 10.0 g of a non-dissociating, nonvolatile solute in 78.11 g of benzene at that temperature. The vapor pressure of the solution was found to be 0.900 atm. Assuming that the solution behaves ideally, determine the molar mass of the solute.

Short answer

The molar mass of the solute which is nonvolatile and non-dissociating is 302.5 g/mol

Step-by-step solution

Determine number of moles

According to the question, it is given that:

The vapor pressure of pure solvent(benzene) = 0.930 atm

The vapor pressure of solution = 0.900 atm

Mass of benzene = 78.11g

Mass of solute = 10.0g

The molar mass of benzene = 78g

We can determine the number of moles of benzene and solute i.e.

Number of moles of benzene = _{$\frac{\text{Mass}}{\text{Molarmass}}\text{=}\frac{78.11}{78}\text{=1mol}$}

Number of moles of solute = _{$\frac{\text{Mass}}{\text{Molarmass}}\text{=}\frac{10}{x}$}

Determine molar mass

According to the relation given by Raoult’s law:

$\begin{array}{c}\frac{\text{P}°\text{-P}}{\text{P}°}\text{=}\frac{\text{Numberofmolesofsolute}}{\text{Numberofmolesofsolute+Numberofmolesofsolvent}}\\ \frac{\text{0.930-0.900}}{0.930}\text{=}\frac{10/\text{x}}{\text{10/x+1}}\\ 0.032\text{=}\frac{10}{10\text{+x}}\\ \left(0.032\right)10\text{+0.032x=10}\\ \text{0.032x=10-0.32}\\ \text{x=302.5g/mol}\end{array}$