Question

The boiling point of a glucose solution containing 12 g of glucose in 100 g of water is 100.34°C. Calculate the molal elevation constant of water. Boiling point of water is 100°C.

Answer 1

Given: Mass of glucose (solute) = 12 g

Molar mass of glucose = 180 g/mol

Mass of water (solvent) = 100 g = 0.1 kg

Boiling point of solution = 100.34°C

Boiling point of pure water = 100°C

Elevation in boiling point ΔT_b = 100.34 − 100 = 0.34°C

Moles of glucose = `12/180` = 0.0667 mol

Molality (m) = `"Moles of solute"/"kg of solvent"`

= `0.0667/0.1`

= 0.667 mol/kg

ΔT_b = K_b . m

`K_b = (Delta T_b)/m`

= `0.34/0.667`

= 0.51°C kg/mol