Question

At a certain temperature, the vapor pressure of pure benzene (C6H6) is 0.930 atm. A solution was prepared by dissolving 10.0 g of a nondissociating, nonvolatile solute in 78.11g of benzene at that temperature. The vapor pressure of the solution was found to be 0.900 atm. Assuming the solution behaves ideally, determine the molar mass of the solute.

Answer 1

Answer: The molar mass of the solute is 300 g/mol

Explanation:

As the relative lowering of vapor pressure is directly proportional to the amount of dissolved solute.

The formula for relative lowering of vapor pressure will be,

$\frac{p^o-p_s}{p^o}=i\times x_2$

where,

$\frac{p^o-p_s}{p^o}$= relative lowering in vapor pressure

i = Van'T Hoff factor = 1 (for non electrolytes)

$x_2$ = mole fraction of solute  =$\frac{\text {moles of solute}}{\text {total moles}}$

Given : 10.0 g of non volatile solute is present in 78.11 g of solvent benzene.

moles of solute = $\frac{\text{Given mass}}{\text {Molar mass}}=\frac{10.0g}{Mg/mol}$

moles of solvent (benzene) = $\frac{\text{Given mass}}{\text {Molar mass}}=\frac{78.11g}{78.11g/mol}=1mole$

Total moles = moles of solute (glycerol)  + moles of solvent (water) = $\frac{10.0}{M}+1$

$x_2$ = mole fraction of solute  =$\frac{\frac{10.0}{M}}{\frac{10.0}{M}+1}$

$\frac{0.930-0.900}{0.930}=1\times \frac{\frac{10.0}{M}}{\frac{10.0}{M}+1}$

$M=300g/mol$

Thus the molar mass of the solute is 300 g/mol