Question

Boiling point of a solvent is 80.2°C. When 0.419 g of the solute of molar mass 252.4 g mol^-1 is dissolved in 75 g of the above solvent, the boiling point of the solution is found to be 80.256 °C. Find the molal elevation constant.

Answer 1

Given : T₀ = (273 + 80.2) K

T_b = 273 + 80.256 (K)

W₁ = 75g 

W₂ = 0.419 g 

M₂ = 252.4 g mol 

K_b = ?

ΔT_b = T_b - T₀

= (273 + 80.256) – (273 + 80.2) 

= 0.056 K

∴ K_b = 2.53 K kg mol^-1