Question

Ethylene dibromide $({C_2}{H_4}B{r_2})$ and $1,2 - \operatorname{di} bromopropane({C_3}{H_6}B{r_2})$ forms a series of ideal solution over the whole range of composition. At ${85^\circ }C$, the vapour pressure of these pure liquids are $183mmHg$ and $127mmHg$ respectively. $10gm$ of ethylene dibromide is dissolved in $80gm$ of $1,2 - \operatorname{di} bromo - propane$.
Calculate the partial pressure of each component 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

Hint: As we know that the ideal solutions are those which obey Raoult’s law over the entire range of concentrations and we are also aware with the Raoult’s law which states that for a solution of volatile liquids, the partial vapour pressure of each component in the solution is directly proportional to its mole fraction.

Formula used: $p = {p^\circ }\chi $, ${p_A} = \chi _A'{P_T}$ and ${P_T} = {p_A} + {p_B}$.

Complete step-by-step answer:
As we have already discussed that Raoult’s law is expressed as shown below:
$p = {p^\circ }\chi $, where$p$ is the partial vapour pressure, $\chi $ is the mole fraction of any component and ${p^\circ }$is the vapour pressure of the pure component.
Now, let us consider $({C_2}{H_4}B{r_2})$as component A and $({C_3}{H_6}B{r_2})$ as component B.
Now, we will first calculate the number of moles of each component so as to find out the mole fraction of each component. Thus, for component A, the mass is $10gm$, and we also know the molecular mass of $({C_2}{H_4}B{r_2})$ is $188g$, so the number of moles of ethylene dibromide would be:
$\Rightarrow moles = \dfrac{{mass}}{{molecular{\text{ }}mass}}$
$\Rightarrow moles = \dfrac{{10}}{{188}} = 0.053mol$
Similarly, for component B, the mass is given as $80gm$ and the molecular mass of $({C_3}{H_6}B{r_2})$ is $202g$, so the number of moles would be:
$moles = \dfrac{{80}}{{202}} = 0.396mol$
Now, we know that the mole fraction is given by the ratio of the moles of a component to the total number of moles in the solution. So, for component A, the mole fraction would be:
\[\Rightarrow {\chi _A} = \dfrac{{0.053}}{{0.053 + 0.396}}\]
$\Rightarrow {\chi _A} = 0.118$
Similarly, for component B, the mole fraction would be:
$\Rightarrow {\chi _B} = \dfrac{{0.396}}{{0.053 + 0.396}}$
$\Rightarrow {\chi _B} = 0.882$
Now we can calculate the partial vapour pressure of each component by putting the calculated values in the formula and we are given with the vapour pressure of both the components.
For component A, the vapour pressure is $183mmHg$and the mole fraction we calculated is ${\chi _A} = 0.118$, so we will get:
$\Rightarrow {p_A} = {\chi _A}p_A^\circ $
$\Rightarrow {p_A} = 0.118 \times 183$
$\Rightarrow {p_A} = 21.594mmHg$
Similarly, for component B, the vapour pressure is $127mmHg$ and the mole fraction is calculated as ${\chi _B} = 0.882$ , so the partial pressure would be:
$\Rightarrow {p_B} = {\chi _B}p_B^\circ $
$\Rightarrow {p_B} = 0.882 \times 127$
$\Rightarrow {p_B} = 112.014mmHg$
And according to Dalton’s law of partial pressure, we know that the total pressure of a solution is given by the sum of the partial pressure of each component present in the solution and is expressed as:
$\Rightarrow {P_T} = {p_A} + {p_B}$
$\Rightarrow {P_T} = 21.594 + 112.014$
$\Rightarrow {P_T} = 133.608mmHg$
Now, we know that the composition of vapour phase in equilibrium with the solution is determined by the partial pressure of the components and the mole fraction. And for component A, it is expressed as:
$\Rightarrow {p_A} = \chi _A'{P_T}$
$\Rightarrow \chi _A' = \dfrac{{21.594}}{{133.608}} = 0.161$
And for component B, the mole fraction can be find our bye simply subtracting the mole fraction of component A from $1$ as $\chi _A' + \chi _B' = 1$
$\chi _B' = 1 - 0.161 = 0.839$

Note: Always remember that the total vapour pressure over the solution is related to the mole fraction of any one component and it varies linearly with the mole fraction of the second component. Also, the total vapour pressure increases or decreases according to the increase in the mole fraction of the second component.