Asked by  | 29th Feb, 2008, 11:30: PM

Expert Answer:


Heptance and octane form an ideal solution at 373 K, The vapour pressures of the pure liquids at this terperature

are 105.2 KPa and 46.8 KPa respectively. If the solution contains 25g of heptance and 28.5g of octane, calculate

(i) vapour pressure exerted by heptane

(ii) vapour pressure exerted by solution

(iii) mole fraction of octane in the vapour phase

apply raoults law p1=105.2 X mole fraction of heptane

p2 = 46.8 X mole fraction of octane solution= p1 +p2

Answered by  | 2nd Mar, 2008, 06:43: PM

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