Ethylene dibromide $(C_{2} H_{4} B r_{2})$ and $1, 2 - d i b r o m o p r o p a n e (C_{3} H_{6} B r_{2})$ forms a series of ideal solution over the whole range of composition. At $85^{\circ} C$ , the vapour pressure of these pure liquids are are $183 m m H g$ and $127 m m H g$ respectively.
$10 g m$ of ethylene dibromide is dissolved in $80 g m$ of $1, 2 - d i b r o m o - p r o p a n e$ . Calculate the partial pressures of each components and the total pressure of the solution at $85^{\circ} C$ .
Calculate the composition of vapour in equilibrium with the above and express as mole fraction of ethylene dibromide.

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Solution

$A : C_{2} H_{4} B r_{2}, B : C_{3} H_{6} B r_{2}$

$n_{A} = \frac{10}{188} = 0.053, n_{B} = \frac{80}{202} = 0.396$

$x_{A} = 0.118, x_{B} = 0.882$

$P_{A} = x_{A} P_{A}^{o} = 183 \times 0.118 = 21.594$ $m m$ $H g$

$P_{B} = x_{B} P_{B}^{o} = 0.882 \times 127 = 112.014$ $m m$ $H g$

Total pressure of solution $= 21.594 + 112.014 = 133.608$ $m m$ $H g$

$x_{A}^{'} =$ Mole fraction of $C_{2} H_{4} B r_{2}$ in vapour $= \frac{21.594}{133.608} = 0.161$

$x_{B}^{'} =$ Mole fraction of $C_{3} H_{6} B r_{2}$ in vapour $= 1 - 0.161 = 0.839$

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