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AP EAMCET · Chemistry · Solutions

At \(298 \mathrm{~K}\), vapour pressures of two pure liquids \(A\) and \(B\) are 200 and \(400 \mathrm{~mm} \mathrm{Hg}\) respectively, if mole fractions of \(A\) and \(B\) in solution are 0.7 and 0.3 respectively? What is the mole fraction of \(B\) in vapour phase?

  1. A 0.279
  2. B 0.721
  3. C 0.538
  4. D 0.462
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Answer & Solution

Correct Answer

(D) 0.462

Step-by-step Solution

Detailed explanation

Given, \[ \begin{aligned} p_A^{\circ} & =200 \mathrm{~mm} \mathrm{Hg} \\ p_B^{\circ} & =400 \mathrm{~mm} \mathrm{Hg} \end{aligned} \] Mole fraction of \(A\) in solution \(=\chi_A=0.7\) Mole fraction of \(B\) in solution \(=\chi_B=0.3\)…
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