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

At T (K), the vapour pressures of pure liquids A and B are 100 mm and

1.	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
Answer» 0.706, 0.294 0.294, 0.706 0.40, 0.60 0.60, 0.40 Solution :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`