18 g of glucose, ${\mathrm{C}}_6{\mathrm{H}}_{12}{\mathrm{O}}_6$ (Molar mass – 180 g mol-1) is dissolved in 1 kg of water in a sauce pan. At what temperature will this solution boil? (Kb for water = 0.52 K kg mol-1, boiling point of pure water = 373.15 K)

We know that : Elevation of boiling point ${\mathrm{\Delta T}}_{\mathrm{b}}$
${\text{ W}}_{\text{B}}{\text{M}}_{\text{B}}\times 100\times \text{ Kbwt. }$of solvent.
Given: WB = 18 g 
MB = Formula of glucose is C6H12O6 
= 6 × 12 + 12 + 6 × 16 = 180 
Wt. of solvent = 1 kg or 1000 g, 
K_b = 0.52 K kg mol-1 
Hence, ΔTb = 18g180×1000×0.521000g = 0.52 K 
∴B.P of the solution = 373.15 + 0.052 
= 373.202 K