Question

If two substance $A$ and $B$ have $P_{A^{o}}: P_{B^{o}}=1: 2$ and have mole fraction in solution $1: 2$ then mol fraction of $A$ in vapours:

• A. 0.33
• B. 0.25
• C. 0.52
• D. 0.2

Answer 1

Answer: (D)

For an ideal solution of component $A$ and $B$ Mole fraction of component $A$ in vapour phase $=\frac{\text { partial pressure of A }}{\text { total vapour pressure }}=\frac{P_{A}}{P_{A}+P_{B}}$
but $P_{A}=P_{A}^{o} x_{A}$ and $P_{B}=P_{B}^{o} x_{B}$
$=\frac{P_{A}^{o} x_{A}}{A_{A}^{o} x_{A}+P_{B}^{o} x_{B}}$
Given: $P_{A}^{o}=1, x_{A}=1 P_{B}^{o}=2, x_{B}=2$
On substituting the values, we get Mole fraction of component $A$ in vapour phase
$=\frac{1 \times 1}{1 \times 1+2 \times 2}=\frac{1}{1+4}$
$=\frac{1}{5}=0.2$