At what temperature a solution containing 5.60 g of glucose per 1000 g of water would boil at 1.013 bar? (Kb for water is 0.52 K kg mol–1).
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 373.166 K
Pure water boils at 373.15 K at 1.013 bar. Glucose is a non-volatile non-electrolyte, so the elevation of boiling point is
\( \Delta T_b = K_b \times m \)
Molar mass of glucose (\( \ce{C6H12O6} \)) = 180 g mol−1:
\( \displaystyle m = \frac{5.60/180\ \text{mol}}{1\ \text{kg}} = 0.0311\ \text{mol kg}^{-1} \)
\( \Delta T_b = 0.52\ \text{K kg mol}^{-1} \times 0.0311\ \text{mol kg}^{-1} = 0.016\ \text{K} \)
\( T_b = 373.15 + 0.016 = 373.166\ \text{K} \)
Note: NTA picked this question from an NCERT (2026-27) example and changed the figures. Do not ignore NCERT while practising numericals.
NCERT (2026-27 edition):
Elevation of Boiling Point – “… 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.”