Question

Explain the reason for depression in vapour pressure. At any temperature, the vapour pressure of pure solvent is 0.745 bar. \( 0.5 \, \text{g} \) non-volatile solute was dissolved in \( 39.0 \, \text{g} \) solvent (molar mass \( 78 \, \text{g mol}^{-1} \)). The vapour pressure of the solution so obtained is 0.740 bar. Find out the molar mass of the solid.

Hint:

Depression in vapour pressure helps determine molar mass of non-volatile solutes.

Answer 1

Depression in Vapour Pressure: \[ \Delta P = P^0 - P = 0.745 - 0.740 = 0.005 \, \text{bar}. \] Relative lowering of vapour pressure: \[ \frac{\Delta P}{P^0} = \frac{n_2}{n_1} = \frac{w_2}{M_2} \cdot \frac{M_1}{w_1}. \] Given: \[ w_2 = 0.5 \, \text{g}, \, w_1 = 39.0 \, \text{g}, \, M_1 = 78 \, \text{g mol}^{-1}. \] Substitute values: \[ \frac{0.005}{0.745} = \frac{0.5}{M_2} \cdot \frac{78}{39}. \] Simplify: \[ M_2 = \frac{0.5 \cdot 78}{0.005 \cdot 39 \cdot 0.745} = 261.86 \, \text{g mol}^{-1}. \]