If 18.0 g of glucose is dissolved in 1 kg of water, boiling point of this solution should be:

Correct answer: C. 100.052 °C

A. 100.52 °C
B. 100.00 °C
C. 100.052 °C
D. Less than 100 °C

Explanation

Mass of glucose = 18 g Mass of solvent = 1 kg Boiling point of pure water = 100°C n (glucose) = 18 / [6(12) + 12(1) + 6(18)] = 18 /180 = 0.1 moles Molality (glucose) = n / mass of solvent = 0.1 / 1 = 0.1 molal ∆Tb = kb × molality = 0.52 × 0.1 = 0.052 Kelvin New boiling point = Boiling point of pure water + ∆Tb New boiling point = 100 + 0.052 = 100.052°C

Last updated 1 September 2026

If 18.0 g of glucose is dissolved in 1 kg of water, boiling point of this solution should be:

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