A solute of molar mass $125 g mo l^{- 1}$ ( $2.5 g$ ) is dissolved in $50 g$ of a solvent. The boiling point of the solvent rises by $0.86 K$ . Calculate $K_{b}$ of the solvent in $K kg mo l^{- 1}$ , correct to two decimal places (solute is a non-electrolyte).

Numerical answer type, answer in K kg/mol, tolerance 0.01.

Answer: 2.15 K kg/mol

Molality

= (2.5/125) /0.050 = 0.40 mol k g^{- 1}

K_{b} = Δ T_{b} / m = 0.86/0.40 = 2.15 K kg mo l^{- 1}