Question

A solution contains equal amounts (in moles) of liquid components A and B. The vapor pressure of pure A is 100 mmHg and that of pure B is 200 mmHg.

What is the vapor pressure above the solution assuming it is an ideal solution?

The experimentally measured vapor pressure is 150 mmHg. What does this indicate about relative strengths of the interactions between solute and solvent molecules?  

Please help me

Answer 1

Let us take A as ethanol, C₂H₆O and B as cyclohexanone, C₆H₁₀O having 4.88 g of ethanol and 8.34 g of cyclohexanone.

It is given that the vapor pressure of pure A is 100 mmHg and that of pure B is 200 mmHg.

Vapor pressure of solution = vapor pressure of pure ethanol(mole fraction ethanol) + vapor pressure of cyclohexanone(mole fraction cyclohexanone).
4.88 of C2H6O x 1mole C2H6O / 16 g = 0.305 moles ethanol
8.34 g C6H10O x 1mole C6H10O / 98 grams = 0.085 moles cyclohexanone

mole fraction ethanol = 0.305 / 0.305 + 0.0851 = 0.305 / 0.39 = 0.781
mole fraction cyclohexanone = 0.0851 / 0.39 = 0.218

vapor pressure solution = 100 (0.781) + 200 (0.218) = 78.1 + 43.6 = 121.7 mm Hg