Question

Calculate the (i) elevation in the boiling point and (ii) the boiling point of 0.05 m aqueous solution of glucose. (K_b = 0.52 Km^-1).

Answer 1

Given : Concentration of the solution = m = 0.05 m

Molal elevation constant = K_b = 1.86 K kg mol^-1

Boiling point of pure water = T₀ = 373 K

Elevation in the boiling point = ΔT_b = ?

Boiling point of solution = T_b = ?

(i) ΔT_b = K_b × m 

= 0.52 × 0.05 

= 0.026K

(ii) The elevation in the boiling point is given by,

ΔT_b = T_b – T₀

∴ Boiling point of a solution,

T_b = T₀ + ΔT_b

= 373 + 0.026

= 373.026 K

∴ ΔT_b = 0.026 K, T_b = 373.026 K