TS EAMCET · Chemistry · Solutions
Liquids A and B form an ideal solution. The vapour pressure of \(A\) and \(B\) are 50 and \(32 \mathrm{~mm} \mathrm{Hg}\) respectively at \(300 \mathrm{~K}\). One mole of liquid \(\mathrm{A}\) is mixed with 1 mole of liquid B. What is the approximate mole fraction of \(\mathrm{A}\) in vapour phase?
- A 0.39
- B 0.5
- C 0.25
- D 0.61
Answer & Solution
Correct Answer
(D) 0.61
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \text { } \mathrm{P}_{\mathrm{A}}=50 ; \mathrm{P}_{\mathrm{B}}=32 \\ & \mathrm{X}_{\mathrm{A}}=\frac{\mathrm{P}_{\mathrm{A}}}{\mathrm{P}_{\mathrm{A}}+\mathrm{P}_{\mathrm{B}}}=\frac{50}{50+32}=\frac{50}{82}=0.609 \approx 0.61\end{aligned}\)
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