18 g of glucose, C6H12O6, is dissolved in 1 kg of water in a saucepan. At what temperature will water boil at 1.013 bar?
Kb for water is 0.52 K kg mol−1
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 373.202 K
Adding a non-volatile solute raises the boiling point of a solvent (elevation of boiling point, a colligative property):
\( \displaystyle \Delta T_b=K_b\,m \)
Here m is the molality: moles of solute per kilogram of solvent. Pure water boils at 373.15 K at 1.013 bar.
\( \displaystyle \text{Moles of glucose}=\frac{18}{180}=0.1\ \mathrm{mol} \)
\( \displaystyle m=\frac{0.1\ \mathrm{mol}}{1\ \mathrm{kg}}=0.1\ \mathrm{mol\,kg^{-1}} \)
\( \displaystyle \Delta T_b=0.52\times0.1=0.052\ \mathrm{K} \)
\( \displaystyle T_b=373.15+0.052=373.202\ \mathrm{K} \)
Shortcut: A solution always boils above the pure solvent. Only 373.202 K is higher than 373.15 K, so no calculation is really needed.
Note: This question is from NCERT (2026-27). NTA sets numerical questions from NCERT too, so do not ignore NCERT while practising numericals.
NCERT (2026-27 edition):
Elevation of Boiling Point – …
