The vapour pressure of a substance of low volatility can be measured by passing an unreactive gas ov — Solutions and Colligative Properties Chemistry Question
Question
The vapour pressure of a substance of low volatility can be measured by passing an unreactive gas over a sample of the substance and analysing the composition of the resulting gas mixture. When nitrogen was passed over mercury at $23^\circ\text{C}$, the mixture analysed to 0.8 mg of Hg, 50.4 g of nitrogen, at a total pressure of 720 torr. The vapour pressure of mercury at this temperature is (Hg = 200)
💡 Solution & Explanation
Moles of Hg $= 0.8 \times 10^{-3} \text{ g} / 200 \text{ g/mol} = 4 \times 10^{-6} \text{ mol}$. Moles of $N_2 = 50.4 \text{ g} / 28 \text{ g/mol} = 1.8 \text{ mol}$. Vapour pressure of Hg = Mole fraction of Hg $\times P_{total} = [4 \times 10^{-6} / (1.8 + 4 \times 10^{-6})] \times 720 \approx (4 \times 10^{-6} / 1.8) \times 720 = 1.6 \times 10^{-3} \text{ torr}$. Therefore, correct answer is B.