The vapour pressure of pure benzene and methyl benzene at 27°C is given as 80 Torr and 24 Torr, respectively.

Question

The vapour pressure of pure benzene and methyl benzene at \(27^\circ C\) is given as 80 Torr and 24 Torr, respectively. The mole fraction of methyl benzene in vapour phase, in equilibrium with an equimolar mixture of those two liquids (ideal solution) at the same temperature is ___ \(\times 10 ^{-2}\) (nearest integer)

Answer 1

Correct answer is : 23

\(Total (P_r) = p ^0 _ {Benzene}\,X_{Benzene} + P^0 _{M.B}X_{M.B}\)

\(P_T = 80 \times \frac {1}{2} + 24 \times \frac {1}{2}\)

In vapor phase = 40 + 12 = 52

Methybemace \(= \frac {P_T}{P}\)

\(=\frac {P^\circ _{M.B}X_{M.B}}{P_T}\)

\(\frac {24\times \frac {1}{2}}{52} = 0.23 \, torr\)

That can be \(=23 \times 10 ^{-2} \,torr\)