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A 0.2 percent aqueous solution of a non-volatile solute exerts vapour pressure of $1.004$ bar at $100^{\large\circ}$C.What is the molar mass of the solute?

$\begin{array}{1 1}4.06g mol^{-1}\\3.06g mol^{-1}\\2.06g mol^{-1}\\1.06g mol^{-1}\end{array} $

1 Answer

According to Raoult's law,
$\Rightarrow \large\frac{n_B}{n_A}$
$\Rightarrow \large\frac{W_B\times M_A}{M_B\times W_A}$
(For dilute solutions,$n_B < < < n_A$)
$M_B=\large\frac{W_B\times M_A}{W_A}.\frac{P^0}{P^0-P_s}$
$\Rightarrow \large\frac{0.2\times 18}{99.8}\times \frac{1.013}{1.013-1.009}$
$\Rightarrow 4.06g mol^{-1}$
answered Aug 8, 2014 by sreemathi.v

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