HardChemistryHigh School (11–12)Solutions › Solubility, Vapour Pressure and Raoult's LawNumericalAP JEE Main

At one temperature, pure A has a vapour pressure of 52.0 kPa and pure B has 18.0 kPa…

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At one temperature, pure A has a vapour pressure of $52.0\ \text{kPa}$ and pure B has $18.0\ \text{kPa}$ . A chemist wants an ideal liquid mixture of the two that gives a total pressure of $30.0\ \text{kPa}$ . What mole fraction of A must the liquid have? What is the mole fraction of A in the vapour above it?

Need a reminder? The key ideaShow

An ideal solution obeys Raoult's law at every composition, with $\Delta_{mix}H = 0$ and $\Delta_{mix}V = 0$ . A positive deviation has a pressure above the Raoult line, a weaker A–B attraction, $\Delta_{mix}H > 0$ and $\Delta_{mix}V > 0$ . A negative deviation is the opposite. The deviation is $\dfrac{P_{measured} - P_{Raoult}}{P_{Raoult}} \times 100\ \%$ .

The worked solution

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