18 g of glucose is dissolved in 1 kg
Asked by admin @ in Chemistry viewed by 273 People
18g of glucose is dissolved in 1Kg of water in a saucepan . At what temperature will water boil at 1.013 bar ? Kb for water is 0.52 Kg mol^-1
Answered by admin @
Moles of glucose = 18 g/ 180 g mol–1 = 0.1 mol
No of kg of solvent = 1 kg
therefore, molality of glucose solution = 0.1 mol kg-1
For water, change in boiling point
ÄTb = 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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