Question

At constant temperature, 6.40 g of a substance dissolved in 78 g of benzene decreases the vapor pressure of benzene from 0.125 atm to 0.119 atm. The molar mass of the substance is ____g mol^-1. (round off to one decimal place)
[Given: Mol. wt. of benzene = 78 g mol^-1]

Answer 1

Correct Answer: 126

\[\text{Given: } m_{\text{solute}}=6.40\ \text{g},\ m_{\text{solvent}}=78\ \text{g},\ P_0=0.125\ \text{atm},\ P=0.119\ \text{atm},\ M_{\text{benzene}}=78\ \text{g mol}^{-1}.\]

\[\frac{P}{P_0}=X_{\text{solvent}}=\frac{0.119}{0.125}=0.952.\]

\[\text{Moles of solvent: } n_{\text{solvent}}=\frac{78}{78}=1.000\ \text{mol}.\]

\[X_{\text{solvent}}=\frac{n_{\text{solvent}}}{n_{\text{solvent}}+n_{\text{solute}}}\]

\[0.952=\frac{1}{1+n_{\text{solute}}}\]

\[1=0.952(1+n_{\text{solute}})\]

\[n_{\text{solute}}=\frac{0.048}{0.952}=0.0504201681\ \text{mol}\]

\[M_{\text{solute}}=\frac{6.40}{0.0504201681}=126.9333333\ \text{g mol}^{-1}\]

\[\boxed{126.9\ \text{g mol}^{-1}}\]