Heptane and octone form ideal solution. At 373 K, the vapour pressure of the two liquid components are 105.2 K Pa and 46.8 K Pa respectively. If the solution contains 25 g of heptane and 25 g of octane, calculate : Vapour pressure exerted by hepthne Vapour pressure exerted by octane Vapour pressure exerted bythe solution Mole fraction octane in the vapour phase.

Heptane and octone form ideal solution. At 373 K, the vapour pressure

1.	Heptane and octone form ideal solution. At 373 K, the vapour pressure of the two liquid components are 105.2 K Pa and 46.8 K Pa respectively. If the solution contains 25 g of heptane and 25 g of octane, calculate : Vapour pressure exerted by hepthne Vapour pressure exerted by octane Vapour pressure exerted bythe solution Mole fraction octane in the vapour phase.
Answer» `"No. of moles of heptane" (n_(B))=("Mass of heptane"(C_(7)H_(16)))/("Gram molar mass")=((25g))/((100 g mol^(-1)))=0.25 mol` `"No. of moles of cotanae" (n_(A))=("Mass of heptane"(C_(8)H_(18)))/("Gram molar mass")=((35g))/((114 g mol^(-1)))=0.307 mol` `"Molar fraction of hepthane" (x_(B))=n_(B)/(n_(B)+n_(A))=((0.25 mol))/((0.25 mol+0.307 mol))=0.449` `"Molar fraction of octane" (x_(A))=n_(A)/(n_(B)+n_(A))=((0.307 mol))/((0.25 mol+0.307 mol))=0.551` `"Vapour pressure of pure heptane" (P_(B)^(@))=105.2 kPa` `"Vapour pressure of pure octane" (P_(A)^(@)) = 46.8 kPa` ` (i) "Vapour pressure of heptane" (P_(B)^(@))=P_(B)^(@)x_(B)=(105.2 kPaxx0.449)47.23 kPa` ` (ii) "Vapour pressure of octane" (P_(A))=P_(A)^(@)x_(A)= (47.23+25.79=73.02 kPa)` `(iii) "Total vapour pressure of solution "(P)=P_(A)+P_(B)=(47.23+25.79)=73.02 kPa` (iv)"mole fraction of octane in the vapour phse"= `P_(A)/(P_(A)+P_(B))=((25.79K Pa))/((73.02 K Pa))= 0.353`

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