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: $P_A^o=1,x_A=1P_B^o=2,x_B=2$
On substituting the values, we get Mole fraction of component $A$ in vapour phase
$=\frac{1\times 1}{1\times 1+2\times 2}=\frac{1}{1+4}$
$=\frac{1}{5}=0.2$