Calculate the boiling point of an aqueous solution containing 18 g glucose in 100 g water if the molal elevation constant of water is 0.5\ K kg mol ^ -1 . [Molar mass of glucose = 180\ g mol ^ -1 and boiling point of water = 100\ ^ C ]

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Correct answer: B. 100.5\ ^ C

Moles of glucose = 18 180 = 0.1\ mol Mass of water = 100\ g = 0.1\ kg Molality of the solution, m = 0.1 0.1 = 1\ mol kg ^ -1 Elevation in boiling point, T_b = K_b m T_b = 0.5 1 = 0.5\ ^ C Boiling point of the solution = 100\ ^ C + 0.5\ ^ C = 100.5\ ^ C Answer: 100.5\ ^ C

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