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if ebullioscopic constant of water is 0.52 K kg/mol and 1 kg of water with 18 g of glucose dissolved in it will boil at 


Solution

Hello Student Moles of glucose = 18 g/ 180 g mol–1 = 0.1 mol Number of kilograms of solvent = 1 kg Thus molality of glucose solution = 0.1 mol kg-1 For water, change in boiling point deltaTb = Kb × m = 0.52 K kg mol–1 × 0.1 mol kg–1 = 0.052 K Since 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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