Correct Answer - Option 2 : x
A = 0.28; x
B = 0.72
Concept:
For ideal solution,
\(p = x_A'\;\;p_A^\circ + x_B'\;p_B^\circ \) ----(1)
Calculation:
\(\because \left[ {x_A' = 0.4,\;x_B' = 0.6} \right]\)
\(p_A^\circ = 7 \times {10^3}\;Pa;\;p_B^\circ = 12 \times {10^3}\;Pa\)
On substituting the given values in Eq. (1) we get,
⇒ p = (0.4 × 7 × 103) + (0.6 × 12 × 103)
⇒ p = 10 × 103 Pa = 1 × 104 Pa
In vapour phase,
\(\Rightarrow {x_A} = \frac{{{p_A}}}{p} = \frac{{x_A^{\rm{'}}{\rm{\;p}}{^\circ _{\rm{a}}}}}{p} = \frac{{0.4 \times 7 \times {{10}^3}}}{{1 \times {{10}^4}}} = 0.28\)
∵ [xA + xB = 1]
⇒ xB = 1 – 0.28
∴ xB = 0.72