Question

4. A solution consists of water and chlorobenzene is distilled at external pressure of 740.2 mmHg. When the solution boils at 90.3 °C, the vapor pressure for pure water is 530.1 mmHg. Calculate the percentage of mass of chlorobenzene in the distillate solution.

Answer 1

P_mix = 740.2 mmHg

P_water =530.1mmHg

P_{chlorobenzene} = P_mix - P_water

= 740.2 mmHg - 530.mmHg

= 210.1 mmHg

Rault's law ------->

P^o_A directly proportional to x_A A ---------> chlorobenzene

P^o_B directly proportional to x_B   B ------->water

P^o_A / P^o_B = x_A / x_B

P^o_A / P^o_B = ( n_A / (n_A + n_B)) / ( n_B/ (n_A + n_B))

= ( n_A / n_B)

P^o_A / P^o_B =   ( n_A / n_B) = (W_A / M_A) / (W_B / M_B)

M_A P^o_A / P^o_B M_B =  (W_A / W_B)

=> (112.5 X 210.1) / (18 X 530.1) =  (W_A / W_B)

2.4771 =  (W_A / W_B)

2.4771 W_B =W_A

mass % of chlorobenzene = W_{chlorobenzene} / W_total X 100

= W_A / (W_A +W_B)  X 100

= 2.44171 W_B / (2.44171 W_B + W_B​​​​​​​) X 100

= (2.44171 W_B / 3.44171 W_B ) X 100

=0.7124 X 100

mass % of chlorobenzene = 71.24 %

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