At T (K), the vapour pressures of pure liquids A and B are 100 mm and 160 mm respectively. An ideal solution is formed by mixing 2 moles of A and 3 moles of B at the same temperature. The mole fraction of A and B in the vapour state respectively are

Options

A. 0.706, 0.294
B. 0.294, 0.706
C. 0.40, 0.60
D. 0.60, 0.40

Answer

B

Explanation

Vapour pressure of solution,
`p_("total") = p_(A) + p_(B), chi_(A)p_(A)^(@) + chi_(B)p_(B)^(@) " "[because p_(A) = chi_(A)p_(A)^(@)]`
Also vapour pressure of component 1, `p_(1) = y_(1) p_("total")`
where y, is the mole fraction of component 1 in vapour phase.
Given,
Vapour pressure of pure liquid A, `p_(A)^(@) = 100 mm`
Vapour pressure ofpure liquid B, `p_(B)^(@) = 160 mm`
`:.` Total vapour pressrure solution `= p_(A) + p_(B)`
`p_("total") = chi_(A)p_(A)^(@) + chi_(B)p_(B)^(@) = (2)/(5) xx 100 + (3)/(5) xx 160`
`= 40 + 96 = 136 mm`
Also `p_(A) = y_(A) p_("total")`
where, `y_(A)` is the mole fraction of A.
Mole fraction of A, `y_(A) = (p_(A))/(p_("total")) = (40)/(136) = 0.294`
`:. y_(B) = 1 - y_(A) = 1 - 0.294 = 0.706`