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 molar mass of the solute?
Here,
Vapour pressure of the solution at normal boiling point (p1) = 1.004 bar
Vapour pressure of pure water at normal boiling point (p10) = 1.013 bar
Mass of solute, (w2) = 2 g
Mass of solvent (water), (w1) = 98 g
Molar mass of solvent (water), (M1) = 18 g mol-1
According to Raoult's law,
\(\frac{p_1^0 - p_1}{p_1^0}\) = \(\frac{w_2 \times M_1}{M_2 \times w_1}\)
⇒ \(\frac{1.013-1.004}{1.013}\) = \(\frac{2 \times 18}{M_2 \times 98}\)
⇒ \(\frac{0.009}{1.013}\) = \(\frac{2 \times 18}{M_2 \times 98}\)
⇒ M2 = \(\frac{1.013 \times 2 \times 18}{0.009 \times 98}\)
= 41.35 g mol-1
Hence, the molar mass of the solute is 41.35 g mol-1.