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18 g of glucose, CgHi2Os is dissolved in 1 kg of water in a saucepan. ExampleAt what temperature will water boil at 1.013 bar? K, for water is 0.52K kg mol |
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Answer» Moles of glucose = 18 g/ 180 g mol–1 = 0.1 molNumber of kilograms of solvent = 1 kgThus molality of glucose solution = 0.1 mol kg-1For water, change in boiling pointÄTb = Kb × m = 0.52 K kg mol–1× 0.1 mol kg–1= 0.052 KSince water boils at 373.15 K at 1.013 bar pressure, therefore, the boiling point of solution will be 373.15 + 0.052 = 373.202 K. |
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