Question

For a binary system of ideal volatile liquids 1 and 2 , the following graph is obtained :


Choose the correct option(s):

• A. When mole fraction of component 1 in liquid phase is 0.4 , the total pressure of the solution is $820mmHg$
• B. When mole fraction of component 1 in liquid phase is 0.4 , the total pressure of the solution is $880mmHg$
• C. At a total pressure of $880mmHg$, mole fraction of component 1 in vapour phase is 0.45
• D. At a total pressure of $850mmHg$, mole fraction of component 1 in vapour phase is 0.25

Answer 1

Answer: (B), (C)

At a total pressure of $880mmHg$, mole fraction of component 1 in vapour phase is 0.45

From graph:
$\text{Vapour pressure of pure component 1 is}1000mmHg$

$\text{Vapour pressure of pure component 2 is}800mmHg$

By Raoult's Law :
$P_T=p_1^{\circ }{\chi }_1+p_2^{\circ }{\chi }_2$
where
$P_T$ is the total pressure of the solution.
$p_1^{\circ }$ is the vapour pressure of pure component 1
$p_2^{\circ }$ is the vapour pressure of pure component 2
${\chi }_1$ is the mole fraction of component 1 in liquid phase.
${\chi }_2$ is the mole fraction of component 2 in liquid phase.

At ${\chi }_1=0.4,{\chi }_2=0.6$
$P_T=1000\times 0.4+800\times 0.6P_T=880mmHg$
Option (b) is correct.
Also,
$y_1\times P_T=p_1^{\circ }{\chi }_1$

where,
$y_1=\text{Mole fraction of component 1 in vapour phase}$
When total pressure is
Putting the values when total pressure is $880mmHg$:
$y_1\times 880=1000\times 0.4y_1=0.45$
Option (c) is correct.

For option (d):
By Raoult's Law :
$P_T=p_1^{\circ }{\chi }_1+p_2^{\circ }{\chi }_2$
$850=1000\times {\chi }_1+800\times (1-{\chi }_1)850=200\times {\chi }_1+800{\chi }_1=\frac{50}{200}=0.25$
This is the mole fraction in liquid phase , not in vapour phase.
For vapour phase mole fraction $\left(y_1\right):$
$y_1\times P_T=p_1^{\circ }{\chi }_1$

where,
$y_1=\text{Mole fraction of component 1 in vapour phase}$
$y_1=\frac{1000\times 0.25}{850}=\frac{5}{17}{y}_1=0.29$
Hence option (d) is incorrect.