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Urea forms an ideal solution in water .Determine the vapour pressure of aqueous solution containing 10% by mass of urea of $4^{\large\circ}C$.(Vapour pressure of water,at $40^{\large\circ}C$=55.3 mm Hg)

$\begin{array}{1 1}\text{53.52 mm Hg}\\\text{63.52 mm Hg}\\\text{73.52 mm Hg}\\\text{83.52 mm Hg}\end{array} $

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Mass of urea = 10g
Mass of water = 90g
Moles of urea =$\large\frac{10g}{60gmol^{-1}}=\frac{1}{6}$
Moles of water =$\large\frac{90}{18}$$=5$
Mole fraction of water
$x_A=\large\frac{5}{5+26}=\frac{5}{31}$
Vapour pressure of solution
$P=P_A^{\large\circ}.x_A$
$\;\;\;=55.3\times \large\frac{5}{31}$
P=53.52 mm Hg
answered Aug 8, 2014 by sreemathi.v
 

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