KCET 2025 Chemistry Paper

To find the boiling point of the water solution, we need to calculate the boiling point elevation using the formula:
∆Tb=i⋅Kb⋅m
where:
∆Tb is the boiling point elevation,
i is the van 't Hoff factor, which is 1 for glucose as it does not dissociate in solution Kb is the ebullioscopic constant for water, given as 0.52K‌kg‌mol−1, m is the molality of the solution.
To calculate the molality m, use:
m=‌

‌ moles of solute ‌

‌ kilograms of solvent ‌

For glucose (C6H12O6), with a mass of 180 g :
‌ Moles of glucose ‌=‌

180g

180g∕mol

=1‌mol
The solvent (water) is 1 kg :
m=‌

1‌mol

1‌kg

=1‌mol∕kg
Substituting into the formula for ∆Tb :
∆Tb=1×0.52×1=0.52K
The normal boiling point of pure water is 373.15 K . Therefore, the boiling point of the solution is:
Tb=373.15+0.52=373.67K