The vapour pressure of a liquid A is 40mmHg at 310K.The vap. pressure of this liquid in a solution with liquid B is 32mmHg . What is the mole fraction of A in the sol. if it obeys Raoults law?
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Pure vapor pressure is 40mmHg
Now after mixing it with a liquid B, vapor pressure has become 32mmHg.
32mmHg is the vapor pressure of the liquid A which corresponds to it's partial vapor pressure.
Let pa be the partial vapor pressure, xa be the mole fraction of A in solution and P is the vapor pressure of pure A
pa = P * xa
Thus, xa = 32/40 = 0.8
Now after mixing it with a liquid B, vapor pressure has become 32mmHg.
32mmHg is the vapor pressure of the liquid A which corresponds to it's partial vapor pressure.
Let pa be the partial vapor pressure, xa be the mole fraction of A in solution and P is the vapor pressure of pure A
pa = P * xa
Thus, xa = 32/40 = 0.8
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