Question

Ethylene dibromide $\left(C_2{H}_{4}B{r}_2\right)$ and $1,2-dibromopropane\left(C_3{H}_{6}B{r}_2\right)$ 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 $183mmHg$ and $127mmHg$ respectively.
$10gm$ of ethylene dibromide is dissolved in $80gm$ of $1,2-dibromo-propane$. 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.

Answer 1

$A:C_2{H}_{4}B{r}_2,B:C_3{H}_{6}B{r}_2$

$n_A=\frac{10}{188}=0.053,n_B=\frac{80}{202}=0.396$

$x_A=0.118,x_B=0.882$

$P_A=x_A{P}_A^o=183\times 0.118=21.594$ $mm$ $Hg$

$P_B=x_B{P}_B^o=0.882\times 127=112.014$ $mm$ $Hg$

Total pressure of solution $=21.594+112.014=133.608$ $mm$ $Hg$

$x_A^{\prime }=$ Mole fraction of $C_2{H}_{4}B{r}_2$ in vapour $=\frac{21.594}{133.608}=0.161$

$x_B^{\prime }=$ Mole fraction of $C_3{H}_{6}B{r}_2$ in vapour $=1-0.161=0.839$