JEE Mains · Chemistry · STD 12 -1. Solution and colligative properties
A solution is made by mixing one mole of volatile liquid \(A\) with 3 moles of volatile liquid \(B\). The vapour pressure of pure A is 200 mm Hg and that of the solution is 500 mm Hg. The vapour pressure of pure \(B\) and the least volatile component of the solution, respectively are what?
- A \(1400 \mathrm{~mm} \mathrm{Hg}, \mathrm{A}\)
- B \(1400 \mathrm{~mm} \mathrm{Hg}, \mathrm{B}\)
- C \(600 \mathrm{~mm} \mathrm{Hg}, \mathrm{B}\)
- D \(600 \mathrm{~mm} \mathrm{Hg}, \mathrm{A}\)
Answer & Solution
Correct Answer
(D) \(600 \mathrm{~mm} \mathrm{Hg}, \mathrm{A}\)
Step-by-step Solution
Detailed explanation
\begin{aligned} & \mathrm{P}_{\mathrm{S}}=\mathrm{P}_{\mathrm{A}}^{\mathrm{o}} \cdot \mathrm{X}_{\mathrm{A}}+\mathrm{P}_{\mathrm{B}}^{\mathrm{o}} \cdot \mathrm{X}_{\mathrm{B}} \\ & 500=200 \times \frac{1}{4}+\mathrm{P}_{\mathrm{B}}^{\mathrm{o}} \cdot \frac{3}{4} \\ &…
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