A solution at 20,^oC is composed of 1.5,mol o | vedclass

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(B) Let $A$ be benzene and $B$ be toluene. Given: $P^o_A = 74.7\,torr$,$P^o_B = 22.3\,torr$,$n_A = 1.5\,mol$,$n_B = 3.5\,mol$. Total moles $n_{total}

Vedclass · Accepted Answer

$38.0\,torr$ and $0.589$

Vedclass · Answer

(B) Let $A$ be benzene and $B$ be toluene. Given: $P^o_A = 74.7\,torr$,$P^o_B = 22.3\,torr$,$n_A = 1.5\,mol$,$n_B = 3.5\,mol$. Total moles $n_{total} = 1.5 + 3.5 = 5.0\,mol$. Mole fraction of benzene in liquid phase,$x_A = \frac{1.5}{5.0} = 0.3$. Mole fraction of toluene in liquid phase,$x_B = \frac{3.5}{5.0} = 0.7$. Total vapour pressure of the solution,$P_{total} = P^o_A x_A + P^o_B x_B = (74.7 	imes 0.3) + (22.3 	imes 0.7) = 22.41 + 15.61 = 38.02\,torr \approx 38.0\,torr$. Mole fraction of benzene in vapour phase $(y_A)$ is given by $y_A = \frac{P_A}{P_{total}} = \frac{P^o_A x_A}{P_{total}} = \frac{22.41}{38.02} \approx 0.589$.

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