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Calculate the mass of a non-volatile solute (molar mass 40 g $mol^{–1}$) which should be dissolved in 114 g octane to reduce its vapour pressure to 80%.

$\begin{array}{1 1}10g\\20g\\30g\\40g\end{array} $

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$P_s=80\%$ of $P_0=0.80$
Solute =$\large\frac{W}{40g mol^{-1}}$
Solvent (octane)=$\large\frac{114g}{114g mol^{-1}}$
(molar mass of $C_8H_{18}=114 gmol^{-1}$)
Now $\large\frac{P_0-P_s}{P_0}=$$x_2$
$\therefore \large\frac{P_0-0.80P_0}{P_0}=\frac{\Large\frac{W}{40}}{\Large\frac{W}{40}+1}$
Or $\large\frac{0.8W}{40}$$=0.2$
$\Rightarrow W= 10g$
answered Aug 5, 2014 by sreemathi.v
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