The vapour pressure of ethanol and methanol ate `44.5 mm Hg` and `88.7 mm Hg`, respectively. An ideal solution is formed at the same temperature by mixing `60g` of ethanol and `40g` of methanol. Calculate the total vapour pressure of the solution and the mole fraction of methanol in the vapour.

Answer

66.15 mm Hg, 0.657

Explanation

Moles of ethanol ` = (60)/(46) = 1.304`
Moles of methanol `= (40)/(32) = 1.250`
Mole fraction of ethaol ` = (1.304)/(1.304 + 1.250) = 0.51`
Mole fraction of ethanol ` = 1-0.51 = 0.49`
p(ethanol) `= 44.5 xx 0.51 = 22.69 ` mm Hg .
p(methanol) ` = 88.7 xx 0.49 = 43.46` mm Hg .
Total vapour pressure ` = 22.69 + 43.46`
` = 66.15` mm Hg .
Mole fraction methanol in vapour phase .
`= ("p(methanol)")/("Total vapour pressure ")`
` = (43.46)/(66.15) = 0.657`