A solution containing $10 g$ per $d m^{3}$ of urea (molecular mass $= 60 g m o l^{- 1}$ ) is isotonic with a $5$ solution of a non-volatile solute. Calculate the molecular mass of this non-volatile solute.

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Solution

According to the equation, $10 g$ per $d m^{3}$ of urea is isotonic with $5 %$ solution of non volatile solute. Therefore, Molar conc. of $5 %$ non volatile solute $= \frac{50 g / d m^{3}}{M o l w t . o f n o n v o l a t i l e s o u t e}$
$\frac{50}{m} M$
Since we know that both solutions are isotoic to each other, hence,
$\frac{1}{6} = \frac{5}{m}$ ; where $m = 50 \times 6 = 300 g / m o l$ .

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