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.

Explanation

Mol. Mass of ethyl alcohol `= C_(2)H_(5)OH=46`
No. of moles of ethyl alcohol `= (60)/(46)=1.304`
Mol. Mass of ethyl alcohol `= CH_(3)OH=32`
No. of moles of methyl alcohol `= (40)/(32)=1.25`
`'x_(A)'`, mole fraction of ethyl alcohol `= (1.304)/(1.304 +1.25)=0.5107`
`'x_(B)'`, mole fraction of methyl alcohol `= (1.25)/(1.304 + 1.25)`
`= 0.4893`
Partial pressure of ethyl alcohol `= x_(a).P_(A)^(0)=0.5107xx44.5`
`=22.73` mm Hg
Partial pressure of methyl alcohol `= x_(B).p_(B)^(0)=0.4893xx88.7`
`=43.40` mm Hg
Total vapour pressure of solution `= 22.73 +43.40`
`= 66.13` mm Hg
Mole fraction of methyl alcohol in the vapour
`= ("Partial pressure of "CH_(3)OH)/("Total vapour pressure")=(43.40)/(66.13)`
`=0.6563`.