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Home  >>  JEEMAIN and AIPMT  >>  Chemistry  >>  Solutions
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Mixture of volatile components A and B has total vapour pressure (in torr) P=$250-115X_A$ where $X_A$ is mol fraction of 'A' in mixture.Hence $P_A^0$ and $P_B^0$ respectively are (in torr)

$\begin{array}{1 1}(a)\;250,120\\(b)\;135,250\\(c)\;250,115\\(d)\;115,250\end{array}$

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1 Answer

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Total pressure $P=250-115X_A$
If $X_A=1$, then $P=P_A^0$
$\therefore P_A^0=250-115\times 1$
$\Rightarrow 135$
$P=250-115(1-X_B)$
$\;\;\;=250-115+115X_B$
If $X_B=1$,then $P=P_B^0$
$\therefore P_B^0=250-115+115\times 1$
$\Rightarrow 250$
Hence (b) is the correct answer.
answered Feb 14, 2014 by sreemathi.v
 
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