Question

Two liquids A and B have the same molecular weight and form an ideal solution. The solution has a vapour pressure of 700 torr at ${80}^{\circ }$ C. It is distilled till $\frac{2}{3}$rd of the solution (mole basis) is collected as condensate. The composition of the condensate is $X_A^{\prime }=0.75$ and that of the residue is $X_A^{\prime \prime }=0.30$. If the vapour pressure of the residue at ${80}^{\circ }$C is 600 torr, which of the following option(s) is/are correct.

• A. The composition of the original liquid was $X_A=0.6$
• B. The composition of the original liquid was $X_A=0.4$
• C. $P_0^A=\frac{2500}{3}$torr
• D. The composition of the original liquid was $X_B$ = 0.4

Answer 1

Answer: (A), (C), (D)

The correct options are
A

The composition of the original liquid was $X_A=0.6$

C

$P_0^A=\frac{2500}{3}$torr

D

The composition of the original liquid was $X_B$ = 0.4

$X_A{P}_A^0+X_B{P}_B^0=700$ ....(i)
$X_A^{\prime \prime }{P}_A^0+X_B^{\prime \prime }{P}_B^0=0.30P_A^0+0.70P_B^0=600$ ... (ii)
The key thing here is that after distillation, 2/3 of total moles are left as condensate and 1/3 of the moles are in the residue

So, if the initial moles of A and B are x and y respectively then we can write:
$x=0.75\times \frac{2}{3}(x+y)+0.3\times \frac{1}{3}(x+y)$ & $X_A=\frac{x}{x+y}$ or $X_B=\frac{y}{x+y}$
$x=0.6(x+y)$, so $X_A=0.6,X_B=0.4$ ... (iii)

This gives us another equation:

$0.6P_A^0+0.4P_B^0=700$ ... (iv)

Solving (iv) and (ii) we get:
$P_A=0.6P_A^0=\frac{2500}{3}$ torr
$P_B=0.4P_B^0=500$ torr