Question:

At 300 K, the vapour pressure of toluene and benzene are 3.63 kPa and 9.7 kPa, respectively. What is the composition of vapour in equilibrium with the solution containing 0.4 mole fraction of toluene?

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To find the mole fraction in the vapour phase, divide the partial pressure of each component by the total pressure.
Updated On: Mar 17, 2025
  • 0.40
  • 0.60
  • 0.80
  • 0.20 

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The Correct Option is D

Solution and Explanation

Using Raoult's law, the partial pressures are \( P_{\text{toluene}} = 0.4 \times 3.63 \, \text{kPa} = 1.452 \, \text{kPa} \) and \( P_{\text{benzene}} = 0.6 \times 9.7 \, \text{kPa} = 5.82 \, \text{kPa} \). The total pressure is \( P_{\text{total}} = 1.452 + 5.82 = 7.272 \, \text{kPa} \). The mole fraction of toluene in the vapour phase is \( \frac{1.452}{7.272} \approx 0.20 \).

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