Question

At 25°C a solution consists of 0.450 mole of pentane, C5H12, and 0.250 mole of cyclopentane, C5H10. What is the lowering of the vapor pressure of pentane in thissolution? The vapor pressure of the pure liquids at 25°C are 451 torr for pentane and 321 torr for cyclopentane. Assume that the solution is an ideal solution.


115 torr

138 torr

161 torr

187 torr

206 torr

Answer 1

p = (mole fraction of solvent) (vapour pressur pressure of pure solvent)
mole fraction of pentane= 0.45/(0.45+0.25)
=0.642
p= 0.642*451 torr
=289.92 torr
lowering in vapour pressure= p(pure solvent)- p(solvent)
= 451-289.92 torr
= 161 torr

Answer 2

138 torr

Answer 3

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Answer 4

161torr