Calculate the boiling point elevation when 5.00 g of glucose ( $C_{6} H_{12} O_{6}$ , $M = 180.16 gmo l^{- 1}$ ) is dissolved in 250.0 g of water. The ebullioscopic constant of water is $K_{b} = 0.512 Kkgmo l^{- 1}$ .

medium 2 marks Class 12 · Chemistry · Solutions

Answer

0.0569 K

The correct answer is 0.0569 K — Apply the boiling point elevation formula: $Δ T_{b} = K_{b} c d o t m = 0.512 Kkgmo l^{- 1} \times 0.1111 m = 0.0569 K$ .

Step-by-step solution

Find the moles of glucose: $\frac{5.00 g}{180.16 gmo l ^{- 1}} = 0.02778 mol$ .
Convert the mass of water to kilograms: $250.0 g = 0.250 kg$ .
Calculate the molality: $\frac{0.02778 mol}{0.250 kg} = 0.1111 m$ .
Apply the boiling point elevation formula: $Δ T_{b} = K_{b} c d o t m = 0.512 Kkgmo l^{- 1} \times 0.1111 m = 0.0569 K$ .

Common mistake: Ignoring the mass of solvent in kg

Concept tested: Boiling point elevation

Published 10 September 2026 · Last updated 16 September 2026 · How this solution was prepared and checked

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Calculate the boiling point elevation when 5.00 g of glucose (C6​H12​O6​, M=180.16gmol−1) is dissolved in 250.0 g of water. The ebullioscopic constant of water is Kb​=0.512Kkgmol−1.

Step-by-step solution

Calculate the boiling point elevation when 5.00 g of glucose ( $C_{6} H_{12} O_{6}$ , $M = 180.16 gmo l^{- 1}$ ) is dissolved in 250.0 g of water. The ebullioscopic constant of water is $K_{b} = 0.512 Kkgmo l^{- 1}$ .