18 g of a non-volatile solute A is dissolved in 1 kg of water, the boiling point of water raised to 373.51 K. Given: Kb for water is 0.52 K kg mol-1. Boiling point for water is 373.15 K at 1.013 bar pressure.
The molecular weight of the solid A is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 26.0 g mol−1
The elevation of boiling point is a colligative property: ΔTb = Kbm. Writing the molality from the masses gives the molar mass of the solute:
\( \displaystyle \Delta T_b = \frac{K_b \times 1000 \times w_2}{M_2 \times w_1} \;\Rightarrow\; M_2 = \frac{K_b \times 1000 \times w_2}{\Delta T_b \times w_1} \)
\( \Delta T_b = 373.51 - 373.15 = 0.36\ \mathrm{K} \)
\( \displaystyle M_2 = \frac{0.52 \times 1000 \times 18}{0.36 \times 1000} = 26\ \mathrm{g\,mol^{-1}} \)