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Vapour pressure of pure liquids P and Q ...

Vapour pressure of pure liquids P and Q are 700 and 450 mm Hg respectively at 330K. What is the composition of the liquid mixture at 330 K, if the total vapour pressure is 600 mm Hg?

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To solve the problem of finding the composition of the liquid mixture of pure liquids P and Q at 330 K, given their vapor pressures and the total vapor pressure, we can use Raoult's Law. Here’s a step-by-step solution: ### Step 1: Understand Raoult's Law Raoult's Law states that the total vapor pressure of a solution (P_total) is equal to the sum of the partial vapor pressures of each component in the mixture. The partial vapor pressure of each component is given by the product of its mole fraction in the liquid phase and its vapor pressure as a pure substance. ### Step 2: Set Up the Equation Let: - \( P_0^P = 700 \, \text{mm Hg} \) (vapor pressure of pure liquid P) ...
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