Home › MCQs › EEJ MAIN Chemistry › Question #7522
Back to Questions
EEJ MAIN Chemistry QUESTION #7522
Question 1

The vapour pressure of pure acetone at 20°C is 185 torr. When 1.2 g of a non-volatile solute is dissolved in 100 g of acetone, the vapour pressure drops to 183 torr. Find the molar mass of the solute (g mol$^{-1}$).

  • 32
  • 64
  • 128✔️
  • 488
Correct Answer Explanation

Using Raoult's Law: $\dfrac{P^0 - P}{P^0} = x_{\text{solute}}$

$\dfrac{185 - 183}{185} = \dfrac{1.2/M}{(1.2/M) + (100/58)}$

$\dfrac{2}{185} = x_{\text{solute}}$

Since $1.2/M \ll 100/58$, approximate: $x_{\text{solute}} \approx \dfrac{1.2/M}{100/58} = \dfrac{1.2 \times 58}{100M}$

$\dfrac{2}{185} = \dfrac{69.6}{100M}$

$M = \dfrac{69.6 \times 185}{200} = \dfrac{12876}{200} \approx \mathbf{128\ \text{g mol}^{-1}}$