Arun
Last Activity: 6 Years ago
Dear 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 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.
