Question

0.90 g of a non-electrolyte was dissolved in 87.90 g of benzene. This raised the boiling point of benzene by 0.25°C. If the molecular mass of the non-electrolyte is 102.0 g mol⁻¹, calculate the molal elevation constant for benzene.

Answer 1

Given: Mass of solute = 0.90 g

Molecular mass of solute = 102.0 g/mol

Mass of benzene (solvent) = 87.90 g = 0.0879 kg

Elevation in boiling point (ΔT_b) = 0.25°C

Moles of solute = `0.90/102`

= 0.00882 mol

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

= `0.00882/0.0879`

= 0.1003 mol/kg

ΔT_b = K_b . m

`K_b = (Delta T_b)/m`

= `0.25/0.1003`

= 2.49 K kg/mol

∴ The molal elevation constant K_b​ for benzene is 2.49 K kg/mol.