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Question

A solution of two substances A and B was formed (mole fractions are $x_{A}$ & $x_{B}$ ). It was found that the mole fraction of A in vapour phase ( $y_{A}$ ) was twice of $x_{A}$ . This is only possible if

$P_{o A} < P_{o B}$

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$P_{o A} > P_{o B}$

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$x_{A} = \frac{P_{o A} - 2 P_{o B}}{2 (P_{o A} - P_{o B})}$

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$x_{A}$ can have any value

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Solution

The correct option is C
$x_{A} = \frac{P_{o A} - 2 P_{o B}}{2 (P_{o A} - P_{o B})}$

$P_{t} y_{A} = P_{A} \Rightarrow y_{A} = \frac{P_{A}}{P_{t}} = \frac{P_{o A} x_{A}}{P_{o A} x_{A} + P_{o B} (1 - x_{A})} \Rightarrow 2 x_{A} = \frac{P_{o A} x_{A}}{(P_{o A} - P_{o B}) x_{A} + P_{o B}} (y_{A} = 2 x_{A}) \Rightarrow 2 (P_{o A} - P_{o B}) x_{A} + 2 P_{o B} = P_{o A} \Rightarrow x_{A} = \frac{P_{o A} - 2 P_{o B}}{2 (P_{o A} - P_{o B})}$

Also, more mole fraction in vapour phase clearly means $P_{o A} > P_{o B}$

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A solution of two substances A and B was formed (mole fractions are xA & xB). It was found that the mole fraction of A in vapour phase (yA) was twice of xA. This is only possible if

The correct option is C xA=PoA−2PoB2(PoA−PoB) PtyA=PA⇒yA=PAPt=PoAxAPoAxA+PoB(1−xA)⇒2xA=PoAxA(PoA−PoB)xA+PoB(yA=2xA)⇒2(PoA−PoB)xA+2PoB=PoA⇒xA=PoA−2PoB2(PoA−PoB) Also, more mole fraction in vapour phase clearly means PoA>PoB

A solution of two substances A and B was formed (mole fractions are $x_{A}$ & $x_{B}$ ). It was found that the mole fraction of A in vapour phase ( $y_{A}$ ) was twice of $x_{A}$ . This is only possible if