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A ideal binary mixture of ethanol and water has a mole fraction of ethanol (x_A) of 0.4. The total pressure of the solution at 298 K is 1 atm. The vapour pressure of pure ethanol (P_A\ ^\circ) and pure water (P_B\ ^\circ) at 298 K are 0.98 atm and 0.95 atm, respectively. Assume that the mixture behaves ideally.

What is the partial pressure of ethanol (P_A) in the vapour phase above the solution?


Option: 1

0.392\, \, atm

Option: 2

0.588 \, \, atm

Option: 3

0.392 \, \, bar

Option: 4

0.588 \, \, bar

Answers (1)


In an ideal solution, the partial pressure of a component in the vapour phase above the solution is equal to the product of its mole fraction in the liquid phase and the vapour pressure of the pure component. Therefore, the partial pressure of ethanol P_A can be calculated as:

P_A=x_A P_A^{\circ}=0.4 \times 0.98 \mathrm{~atm}=0.392 \mathrm{~atm}

Therefore, the correct answer is option (1).

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