Calculate the boiling point elevation when 5.00 g of glucose (C${}_6$H${}_{12}$O${}_6$, $M=180.16{\text{g mol}}^{-1}$) is dissolved in 250 g of water. The ebullioscopic constant $K_b$ of water is $0.512{\text{K kg mol}}^{-1}$.

Step-by-step solution

1. Find moles of glucose: $n=\frac{5.00\text{g}}{180.16{\text{g mol}}^{-1}}=0.0277\text{mol}$.
2. Convert the mass of water to kilograms: $250\text{g}=0.250\text{kg}$.
3. Calculate molality: $m=\frac{n}{\text{kg of solvent}}=\frac{0.0277\text{mol}}{0.250\text{kg}}=0.111{\text{mol kg}}^{-1}$.
4. Apply the boiling point elevation formula: $\Delta T_b=K_b\times m=0.512{\text{K kg mol}}^{-1}\times 0.111{\text{mol kg}}^{-1}=0.0569\text{K}$.
5. Round to three significant figures: $\Delta T_b\approx 0.057\text{K}$.

Concept tested: Boiling point elevation (colligative property)

Published 31 August 2026 · How this solution was prepared and checked

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