The molal elevation constant of water is 0.51 K kg mol−1. The boiling point of a 0.1 molal aqueous NaCl solution (assume complete dissociation) is nearly:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 100.1°C
Elevation of boiling point is a colligative property. NaCl dissociates completely, \( \ce{NaCl -> Na^+ + Cl^-} \), so each formula unit gives two particles and the van’t Hoff factor i = 2:
\( \displaystyle \Delta T_b=i\,K_b\,m \)
\( \displaystyle \Delta T_b=2\times 0.51\,\mathrm{K\,kg\,mol^{-1}}\times 0.1\,\mathrm{mol\,kg^{-1}} \)
\( \displaystyle \Delta T_b=0.102\,\mathrm{K} \)
A rise of 0.102 K is a rise of 0.102 °C, so the boiling point is 100 + 0.102 = 100.102 °C, that is nearly 100.1 °C.