JEE Mains · Chemistry · STD 12 -1. Solution and colligative properties
The vapour pressures of pure liquids \(A\) and \(B\) are \(400\) and \(600\, mm\, Hg\), respectively at \(298\, K\). On mixing the two liquids, the sum of their initial volumes is equal to the volume of the final mixture. The mole fraction of liquid \(B\) is \(0.5\) in the mixture. The vapour pressure of the final solution, the mole fractions of components \(A\) and \(B\) in vapour phase, respectively are
- A \(500\, mmHg, 0.4, 0.6\)
- B \(500\, mmHg, 0.5, 0.5\)
- C \(450\, mmHg, 0.5, 0.5\)
- D \(450\, mmHg, 0.4, 0.6\)
Answer & Solution
Correct Answer
(A) \(500\, mmHg, 0.4, 0.6\)
Step-by-step Solution
Detailed explanation
\({P_{total}}\, = \,{X_A}\,P_A^o\, + \,{X_B}\,P_B^o\) \(\, = \,0.5\, \times \,400\,\, + \,0.5\, \times \,600\) \( = \,500\,mm\,Hg\) Now , mole fraction of \(A\) in vapour \({Y_A}\, = \,\frac{{{P_A}}}{{{P_{total}}}}\, = \,\frac{{0.5\, \times \,400}}{{500}}\, = \,0.4\) and mole…
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