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Liquids A and B form an ideal solution in the entire composition range. At 350 K, the vapour pressures of pure A and pure B are 7 × 103 Pa and 12 × 103 Pa, respectively. The composition of the vapour in equilibrium with a solution containing 40 mole percent of A at this temperature is
1. xA = 0.76; xB = 0.24
2. xA = 0.28; xB = 0.72
3. xA = 0.4; xB = 0.6
4. xA = 0.37; xB = 0.63

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Correct Answer - Option 2 : xA = 0.28; xB = 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

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