Methyl alcohol and ethyl alcohol have vapour pressure equal to `88.5` mm and `42.0` mm respectively `45^(@)`C. If `16.0` g of methanol and `46.0` g ethanol are mixed at `45^(@)`C, the mole fraction of methanol in the vapour is …….(the mixture may be taken as ideal solution).

Options

A. 0.467
B. 0.502
C. 0.513
D. 0.556

Answer

C

Explanation

For and ideal solution, obeying Raoult's law, the vapour pressure of mixture
`"P"_("mixture")=underset("methanol")"P"^(@).underset("methanol")chi^(@)+underset("ethanol")"P"^(@).underset("ethanol")chi^(@)`
Or `"P"_("mixture")=88.5xx(16//32)/(16/32+46/46)+42.0xx(46//46)/(16/32+46/46)`
`"P"_("mixture")=88.5xx0.5/1.5+42.0xx1/1.5=57.44"mm".`
The mole fraction of any constituent 'i' in vapour phase is
`="P"_("i")/("P"_("total")) " that is ""vap. pressure of the constituent in solution"/("Total vap. pressure of solution")`
`=chi_(i) P_(i)^(**)//"P"_("total")=88.5xx[(16//32)/(16//32+46//46)]//57.44`
Or the mole fraction of methanol in vapour phase `=29.47//57.44=0.5130`