Question

An ideal solution contains two volatile liquids $A(p=100torr)$ and $B(p=200torr)$. If mixture contain 1 mole of $A$ and 4 mole of $B$ then total vapour pressure of the distillate is:

• A. $150$
• B. $180$
• C. $188.88$
• D. $198.88$

Answer 1

The correct option is B $180$

Solution:- (B) $180$

Given:-

No. of moles of A $\left(n_A\right)=1\text{mole}$

No. of moles of B $\left(n_B\right)=4\text{moles}$

As we know that,

$\text{mole fraction}=\frac{\text{no. of moles}}{\text{total moles}}$

Therefore,

$x_A=\frac{1}{1+4}=0.2$

$x_B=\frac{4}{1+4}=0.8$

Again, as we know that,

$P_T=P_A+P_B$

$\Rightarrow P_T=(x_A\times {p^{\circ }}_A)+(x_B\times {p^{\circ }}_B)$

Given:-

${p^{\circ }}_A=100\text{torr.}$

${p^{\circ }}_B=200\text{torr.}$

$\therefore P_T=(0.2\times 100)+(0.8\times 200)$

$\Rightarrow P_T=20+160=180\text{torr.}$

Hence the total vapour pressure of the distillate is $180\text{torr.}$.