The freezing point of an aqueous solution is −0.186°C. What is the elevation in the boiling point of the same solution? (Kf = 1.86 K kg mol−1 and Kb = 0.512 K kg mol−1)
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 0.0512 K
Both colligative changes depend on the same molality m of the solution: \( \Delta T_f=K_f\,m \) and \( \Delta T_b=K_b\,m \). Water freezes at 0°C, so the depression is 0.186 K.
Molality from the freezing point:
\( \displaystyle m=\frac{\Delta T_f}{K_f} \)
\( \displaystyle m=\frac{0.186\,\mathrm{K}}{1.86\,\mathrm{K\,kg\,mol^{-1}}}=0.1\,\mathrm{mol\,kg^{-1}} \)
Elevation of boiling point of the same solution:
\( \displaystyle \Delta T_b=K_b\,m \)
\( \displaystyle \Delta T_b=0.512\,\mathrm{K\,kg\,mol^{-1}}\times0.1\,\mathrm{mol\,kg^{-1}} \)
\( \displaystyle \Delta T_b=0.0512\,\mathrm{K} \)