18 g of glucose (molar mass 180 g/mol) is dissolved in 1 kg of water. Given Kb for water is 0.52 K kg/mol and water boils at 373.15 K, the boiling point of the solution is:

1. Moles of glucose = $18 / 180 = 0.1$ mol; molality = $0.1 / 1 = 0.1$ m. 2. $\Delta T_b = K_b m = 0.52 \times 0.1 = 0.052$ K. 3. Boiling point = $373.15 + 0.052 = 373.202$ K. (Option A ignores the elevation.) _Source: NCERT Class 12 Chemistry Ch 1 "Solutions", p.16_