Explanation Vapour pressure of heptane, P heptane o =428mm Hg Vapour pressure of octane, P octane o =175mm Hg Vapour pressure of the solution of the heptane and octane mixture, P solution =305mm Hg Now, let us suppose that one mole fraction of the heptane be X h e p t a n e and the mole fraction of the octane be X o c t a n e . Sum of the mole fractions of the two components of a solution is a utility. X heptane +X octane =1 X octane =1-X heptane So, According to Raoult’s law for a mixture of liquids, the vapour pressure of a solution containing two volatile liquids is equal to the sum of the vapour pressures of the pure liquids calculated by the Raoult’s law. Thus, Pressure of the solution, P solution = ( P heptane o ×X heptane ) + ( P octane o ×X octane ) = (

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Chapter 12, Problem 12.108SP

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Mole fraction of heptane

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Chapter 12, Problem 12.107SP

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Mole fraction of heptane

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Mole fraction of heptane