Question

Calculate ΔT_b and boiling point of 0.05 m aqueous solution of glucose. (Given: K_b = 0.52 K m⁻¹).

Answer 1

Given: m = 0.05 M, K_b= 0 .52 K m⁻¹ ΔT_b = ?

ΔT_b =  K_b × m

= 0.052 × 0.05

= 0.026 K

T_b = Boiling point of solution

`T_b^0` = Boiling point of pure solvent (water)

= 100°C = 273 + 100 = 373 K

We have,

`ΔT_b = T_b - T_b^0`

∴ `T_b = T_b^0 + ΔT_b`

= 373 + 0.026

∴ T_b = 373.026 K