The vapour pressure of ethanol and methanol are `44.0 mm Hg` and `88.0 mm Hg`, respectively. An ideal solution is formed at the same temperature by mixing `60 g` of ethanol with `40g` of methanol. Calculate the total vapour pressure of the solution and the mole fraction of methanol in the vapour.

Answer

0.66

Explanation

Total vapour pressure of solution:
`=(P_(eth)^(@)xxchi_(eth)) + (P_(meth)^(@)xxchi_(meth))`
`=ubrace(P_(eth)^(@)xxchi_(eth))_(P_(eth)) + ubrace(P_(meth)^(@)xxchi_(meth))_(P_(meth))`
Given that:
`p_(eth)^(@)=44.5 mm Hg`,`p_(meth)^(@)=88.7 mm Hg`
Wt. of ethanol =`60 g` , Wt. of methanol =`40 g`
`n_(eth)=60/40=1.304 (:. "molecular weight of ethanol" =40)`
`n_(meth)=40/32=1.25 (:. "molecular weight of methanol" =32)`
`chi_(eth)=(n_(eth))/(n_(eth)+n_(meth))=1.304/(1.304 + 1.25) =(1.304/2.554) =0.51`
`chi_(meth)=(n_(meth))/(n_(meth)+n_(eth))=(1.25/2.554) =0.49`
`:. P_(m)=44.5 xx 0.51 + 88.7 xx 0.49`
`=22.67 +43.46`
=`66.15 mm Hg`
Mole fraction of methanol in vapour =`P_(meth)/P_(m)=(43.46/66.13)=0.66 mm Hg`