Question

The boiling point of pure water is 373 K. Calculate the boiling point of an aqueous solution containing 18 g of glucose (MW = 180) in 100 g of water. Molal elevation constant of water is 0.52 K kg mol⁻¹.

Answer 1

Given: Mass of glucose = 18 g

Molar mass of glucose (MW) = 180 g/mol

Mass of water = 100 g = 0.1 kg

K_b (molal elevation constant for water) = 0.52 K kg mol⁻¹

Boiling point of pure water = 373 K

Moles of glucose = `18/180`

= 0.1 mol

Molality (m) = `(0.1  mol)/(0.1  kg)`

ΔT_b = K_b m

= 0.52 × 1.0

= 0.52 K

`T_"solution" = T_"pure water" + Delta T_b`

= 373 + 0.52

= 373.52 K