The vapour pressure of ethanol and methanol are 44.5 and 88.7 mm of Hg at 298 K. An ideal solution is formed at the same temperature by mixing 60 g of ethanol and 40 g methanol. Calculate the total vapour pressure of the solution and the mole fraction of methanol in the vapour phase.

Explanation

We are given that
Weight of methanol `(w_(B))=40g`
Weight of ethanol `(w_(A))=60g`
Vapour pressure of methanol `(p_(B)^(@))=88.7" mm of Hg"`
Vapour pressure of ethanol `(p_(A)^(@))=44.5" mm of Hg"`
Molecular mass of methanol `(M_(B))=32g` Moles of methanol `(n_(B))=w_(B)/(M_(B))=40/(32)=1.25"mol"`
Molecular mass of ethanol `(M_(A))=46g` Moles of ethanol `(n_(A))=w_(A)/(M_(A))=60/(46)=1.30"mol"`
Mole fraction of methanol `(x_(B))=n_(B)/(n_(A)+n_(B))=1.25/(1.30+1.25)=0.49`
Mole fraction of ethanol `(x_(A))=n_(A)/(n_(A)+n_(B))=1.30/(1.30+1.25)=0.51`
Partial pressure of methanol `(p_(B))=p_(B)^(@)x_(B)=88.7xx0.49=43.46"mm"`
Partial pressure of ethanol `(p_(A))=p_(A)^(@)x_(A)=44.5xx0.51=22.70"mm"`
`thereforeP=p_(A)^(@)x_(A)+p_(B)^(@)x_(B)=22.70+43.46=66.16"mm"`
Therefore, vapour pressure of the solution is 66.16 mm of Hg.