Question

Pressure of pure benzene is 0.75. When 0.2g of a non-volatile solute is added to 39g of benzene, pressure changes to 0.745. The molar mass of solute is?

• A. 40 g/mol
• B. 100 g/mol
• C. 60 g/mol
• D. 80 g/mol

Hint:

To calculate the molar mass of a solute, use Raoult's Law and relate the change in vapor pressure to the mole fraction of the solute.

Answer 1

The Correct Option is C

The change in pressure is related to the molar mass of the solute using **Raoult's Law**, which states that the relative lowering of vapor pressure is proportional to the mole fraction of the solute: \[ \frac{\Delta P}{P_{\text{initial}}} = \frac{n_{\text{solute}}}{n_{\text{solvent}}} \] Where: - \( \Delta P \) is the change in vapor pressure, - \( P_{\text{initial}} \) is the initial vapor pressure, - \( n_{\text{solute}} \) and \( n_{\text{solvent}} \) are the moles of solute and solvent. ### Step 1: Calculate the Change in Vapor Pressure The initial pressure of pure benzene is 0.75, and the final pressure is 0.745. The change in pressure \( \Delta P \) is: \[ \Delta P = P_{\text{initial}} - P_{\text{final}} = 0.75 - 0.745 = 0.005 \] ### Step 2: Apply Raoult's Law The number of moles of solvent \( n_{\text{solvent}} \) (benzene) is: \[ n_{\text{solvent}} = \frac{m_{\text{solvent}}}{M_{\text{solvent}}} = \frac{39 \, \text{g}}{78 \, \text{g/mol}} = 0.5 \, \text{mol} \] The mole fraction of the solute is: \[ \frac{n_{\text{solute}}}{n_{\text{solvent}}} = \frac{\Delta P}{P_{\text{initial}}} = \frac{0.005}{0.75} = 0.00667 \] Thus: \[ n_{\text{solute}} = 0.00667 \times 0.5 = 0.00333 \, \text{mol} \] ### Step 3: Calculate the Molar Mass of the Solute The molar mass of the solute is: \[ M_{\text{solute}} = \frac{m_{\text{solute}}}{n_{\text{solute}}} = \frac{0.2 \, \text{g}}{0.00333 \, \text{mol}} = 60 \, \text{g/mol} \] Thus, the molar mass of the solute is: \[ \boxed{(C) \, 60 \, \text{g/mol}} \]