A mixture of chlorobenzene and water (which are virtually immiscible) boils at `90.3 ^@C` at an external pressure of 740.2 mm. The vapour pressure of pure water at `40.3 ^@C` is 530.1 mm. What is the percentage weight of chlorobenzene in the distillate?

Options

A. 20.21
B. 37.19
C. 28.74
D. 71.26

Answer

D

Explanation

Partial pressure of liquids are proportional to their molar concentrations.
`P_A/P_B = N_A/N_B = (W_A//M_A)/(W_B//M_B)`
`P_A and P_B` are the partial pressures of two immiscible liquids A and B.
Similarly, partial pressure of chlorobenzene,
`P_c = 740.2 - 530.1 = 210.1 mm`
Partial pressure of water `= P_w = 530.1 mm`
`P_c/P_w= N_c/N_w = (W_cxxM_w)/(W_w XXM_c) = (W_c xx18)/(W_wxx112.5)`
`P_c/P_w = (210.1)/53.01 = (W_cxx18)/(W_wxx112.5)implies W_c/W_w =(210.1XX112.5)/(530.1xx18) = 2.48`
`W_c/W_w = 2.48 or W_c = 2.48xxW_w`
or `W_c +W_w = 3.48 xx W_w`
`%W_c = (2.48xxW_w)/(3.48xxW_w)xx100=71.26`