The rise in the boiling point of a solution containing 1.8 g of glucose in 100 g of solvent is 0.1°C. The molal elevation constant of the liquid is:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 1 K kg mol−1
A non-volatile solute raises the boiling point of a solvent. The elevation of boiling point is proportional to the molality m of the solution; the constant is the molal elevation constant Kb:
\( \displaystyle \Delta T_b=K_b\,m \)
Glucose, \( \ce{C6H12O6} \): M = 6(12) + 12(1) + 6(16) = 180 g mol−1. It is a non-electrolyte, so it does not dissociate.
\( \displaystyle n=\frac{1.8\,\mathrm{g}}{180\,\mathrm{g\,mol^{-1}}}=0.01\,\mathrm{mol} \)
\( \displaystyle m=\frac{0.01\,\mathrm{mol}}{0.100\,\mathrm{kg}}=0.1\,\mathrm{mol\,kg^{-1}} \)
A rise of 0.1°C is a rise of 0.1 K, since a temperature difference is the same on both scales.
\( \displaystyle K_b=\frac{\Delta T_b}{m}=\frac{0.1\,\mathrm{K}}{0.1\,\mathrm{mol\,kg^{-1}}} \)
\( \displaystyle K_b=1\,\mathrm{K\,kg\,mol^{-1}} \)
The molal elevation constant of the liquid is 1 K kg mol−1.