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Home  >>  CBSE XII  >>  Chemistry  >>  Solutions
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Two liquids A and B form ideal solution.At 300K,the vapour pressure of solution containing 1 mol of A and 3 mol of B is 550mm Hg.At the same temperature,if one more mole of B is added to this solution,the vapour pressure of the solution increases by 10mm Hg.Determine the vapour pressures of A and B in their pure states.

$\begin{array}{1 1}400mm,600mm\\200mm,400mm\\500mm,200mm\\300mm,800mm\end{array} $

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Since $P=x_AP_A^0+x_BP_B^0$
$\Rightarrow \big(\large\frac{1}{1+3}\big)$$P_A^0+\big(\large\frac{3}{1+3}\big)$$P_B^{\large\circ}$=550mm Hg
$\Rightarrow \big(\large\frac{1}{1+4}\big)$$P_A^0+\big(\large\frac{4}{1+4}\big)$$P_B^{\large\circ}$=560mm Hg
That is,
$0.25P_A^0+0.75P_B^0$=550mm Hg
$0.20P_A^0+0.8P_B^0$=560mm Hg
Solving for $P_A^0$ and $P_B^0$ we get,
$P_A^0=400$ mm Hg
$P_B^0=600$ mm Hg
answered Aug 7, 2014 by sreemathi.v
 
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