SolutionHard
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°C, the mixture was 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)
Options
A.8 × 10−4 torr
B.1.6 × 10−3 torr
C.3.2 × 10−3 torr
D.1.6 × 10−4 torr
Solution
$P_{Hg} = X_{Hg}.P_{\text{total}} = \frac{\frac{0.8 \times 10^{- 3}}{200}}{\frac{0.8 \times 10^{- 3}}{200} + \frac{50.4}{28}} \times 720 = 1.6 \times 10^{- 3}\text{ torr}$
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