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.

Explanation

`"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`