Question

The vapor pressure of toluene (Mol. Wt. = 92) is 0.13 atm at 25 °C. If 6 g of a hydrocarbon is dissolved in 92 g of toluene, the vapor pressure drops to 0.12 atm. The molar mass of the hydrocarbon (rounded off to the nearest integer) is \(\underline{\hspace{2cm}}\).

Hint:

Raoult's Law can be used to determine the mole fraction of solute and the corresponding change in vapor pressure when a non-volatile solute is dissolved in a solvent.

Answer 1

Correct Answer: 71

We can use Raoult's Law to solve this problem, which relates the change in vapor pressure to the mole fraction of the solute. The formula is:
\[ \Delta P = P_0 \times X_{\text{solute}} \] where \( \Delta P \) is the change in vapor pressure, \( P_0 \) is the initial vapor pressure, and \( X_{\text{solute}} \) is the mole fraction of the solute. The mole fraction of the solute is given by:
\[ X_{\text{solute}} = \frac{n_{\text{solute}}}{n_{\text{solute}} + n_{\text{solvent}}} \] We can find \( n_{\text{solute}} \) using the mass and molar mass of the hydrocarbon:
\[ n_{\text{solute}} = \frac{6}{M} \] where \( M \) is the molar mass of the hydrocarbon. The mole fraction and Raoult's law give:
\[ 0.01 = \frac{6}{M} \div \left( \frac{6}{M} + \frac{92}{92} \right) \] Solving for \( M \), we find:
\[ M \approx 73 \, \text{g/mol}. \] Thus, the molar mass of the hydrocarbon is \( 73 \, \text{g/mol} \).