A solution boils at 100.26°C at 1 atm. What is the molality if Kb = 0.52 K kg mol⁻¹ ?
Δ Tb = Kb · m . 100.26 - 100 = 0.52 · m . m = (0.26/0.52) = 0.5 mol/kg .
Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Solubility - of Solids and Gases in Liquids Henry's Law