Vapor pressure of A `("molar mass" =78 mol^(-1))` and `B ("molar mass" =92 g mol^(-1))` at `20^(@)C` are `75 mmHg` and `22 mmHg` respectively. `23.4 g` of `A` and `64.4 g` of `B` are mixed to form an odeal solution. If the vapors are in equilibrium with the liquid mixture at the same tamperature, then the mole fraction of `A` in the vapor phase will be

Options

A. `0.59`
B. `0.31`
C. `0.69`
D. `0.41`

Answer

1

Explanation

According to Dalton's Law
`chi_(A)=P_(A)/P_("total")`
`=(P_(A)^(0)chi_(A))/(P_(A)^(0)chi_(A)+P_(B)^(0)chi_(B)) …(1)`
Thus we have to find the numbers of moles of solvent and of solute and then calculate the mole fractions.
`n_(A)=(mass_(A))/(molar mass_(A))=(23.4 g)/(78 g mol^(-1))=0.3 mol`
`n_(B)=(mass_(B))/(molar mass_(B))=(64.4 g)/(92 g mol^(-1))=0.7 mol`
`:. chi_(A)=0.3/(0.3+0.7)=0.3`
`chi_(B)=1-chi_(A)=0.7`
Substituting these results into Equation (1), we get
`chi_(A)=((75 mmHg)(0.3))/((75 mmHg)(0.3)+(22 mmHg)(0.7))`
`=(22.5 mmHg)/(22.5 mmHg+15.4 mmHg)`
`=(22.5 mmHg)/(37.9 mmHg)`
`=0.59`