Discussion
The vapour pressure of 0.5 M aqueous KNO3 solution is 749.7 torr at 100°C. The activity of water in this solution at 100°C is :
Answer: D
The activity of water is 0.9864.
The activity of the solvent is the ratio of its vapour pressure over the solution to that of the pure solvent at the same temperature. At 100°C pure water boils, so its vapour pressure is 760 torr:
\[ a_{w} = \frac{p}{p^{*}} = \frac{749.7}{760} = 0.9864 \]
The activity of water in a solution must be less than 1, because the dissolved KNO3 lowers the vapour pressure. That rules out 1.0023 and 1.0230, which are above 1. 0.9218 is far too low for a 0.5 M solution, whose vapour pressure falls by only about 10 torr.
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