Question

At ${80}^{°}\mathrm{C}$ the vapour pressure of pure liquid A is $520\mathrm{mm}\mathrm{of}\mathrm{Hg}$ and that of pure liquid B is $1000\mathrm{mm}\mathrm{Hg}$. If a mixture solution of A and B boils at and $1$ atm pressure, the amount of A (mole percent) in the mixture is$(1\mathrm{atm}=760\mathrm{mm}\mathrm{Hg})$.

• A. $50\%$
• B. $54\%$
• C. $32\%$
• D. $44\%$

Answer 1

The correct option is A

$50\%$

Explanation for correct option:

(A) $50\%$

Step 1: Given data and formula:

$$\begin{gathered} {\mathrm{P}}_{\mathrm{A}}^{°}=520\mathrm{mm}\mathrm{Hg} \\ {\mathrm{P}}_{\mathrm{B}}^{°}=1000\mathrm{mm}\mathrm{Hg} \\ {\mathrm{P}}_{\mathrm{T}}=760\mathrm{mm}\mathrm{Hg} \\ {\mathrm{P}}_{\mathrm{T}}={\mathrm{X}}_{\mathrm{A}}{\mathrm{P}}_{\mathrm{A}}^{°}+{\mathrm{X}}_{\mathrm{B}}{\mathrm{P}}_{\mathrm{B}}^{°} \\ \mathrm{Where}, \\ {\mathrm{P}}_{\mathrm{T}}-\mathrm{Total}\mathrm{vapour}\mathrm{pressure} \\ {\mathrm{P}}_{\mathrm{A}}^{°},{\mathrm{P}}_{\mathrm{B}}^{°}-\mathrm{vapour}\mathrm{pressure}\mathrm{of}\mathrm{solution}\mathrm{A}\mathrm{and}\mathrm{B}. \\ {\mathrm{X}}_{\mathrm{A}},{\mathrm{X}}_{\mathrm{B}}-\mathrm{Mole}\mathrm{fraction}\mathrm{of}\mathrm{A}\mathrm{and}\mathrm{B} \end{gathered}$$

Step 2: Substitute the values:

$$\begin{gathered} {\mathrm{P}}_{\mathrm{T}}={\mathrm{X}}_{\mathrm{A}}{\mathrm{P}}_{\mathrm{A}}^{°}+{\mathrm{X}}_{\mathrm{B}}{\mathrm{P}}_{\mathrm{B}}^{°} \\ 760=520{\mathrm{X}}_{\mathrm{A}}+1000(1-{\mathrm{X}}_{\mathrm{A}})(\mathrm{Since}{\mathrm{X}}_{\mathrm{A}}+{\mathrm{X}}_{\mathrm{B}}=1) \end{gathered}$$

Step 3: Finding the value of the mole fraction of B:

$$\begin{gathered} 760=520{\mathrm{X}}_{\mathrm{A}}+1000-1000{\mathrm{X}}_{\mathrm{A}} \\ 240=480{\mathrm{X}}_{\mathrm{A}} \\ {\mathrm{X}}_{\mathrm{A}}=\frac{1}{2} \\ =0.5 \\ \mathrm{To}\mathrm{get}\mathrm{percent}\mathrm{multiply}\mathrm{wth}100 \\ {\mathrm{X}}_{\mathrm{A}}=50\% \end{gathered}$$

Hence, the correct option is (C) $50\%$.