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Q.
An aqueous solution of $2\%$ non-volatile solute exerts a pressure of $1.004$ bar at the normal boiling point of the solvent. What is the molecular mass of the solute?
Solutions
Solution:
Vapour pressure of pure water at boiling point
$=1\, atm = 1.013 \,bar$
Vapour pressure of solution $(ps) = 1.004\, bar$
Let mass of solution $= 100\, g$
Mass of solute $= 2 \,g$
Mass of solvent $= 100 - 2 = 98\, g$
$\frac{p^{\circ}-p_{s}}{p^{\circ}} = \frac{n_{2}}{n_{1}+n_{2}}$
$=\frac{n_{2}}{n_{1}}=\frac{W_{2}/M_{2}}{W_{1}/M_{1}} \left(\because\,n_2 <<< n_{1}\right)$
$\frac{1.013-1.004}{1.013} = \frac{2}{M_{2}} \times \frac{18}{98}$
or $M_{2 } = \frac{2 \times 18}{98} \times \frac{1.013}{0.009}$
$= 41.35\,g\,mol^{-1}$