18 gram Glukosa (Mr)= 180 dilarutkan dalam 500 gr air. Tentukan titik didih dan titik beku larutan (Kb) air= 0,52°C/m, dan kr air- 1,86°C/m)

wt = 18 gram
Mr = 180
wp = 500 gram
Kb = 0,52°C
Kf = 1,86°C
Kb = ...?
Kf = ...?

m = wt/Mr x 1000/wp
= 18/180 x 1000/500
= 0,1 x 2
= 0,2 molal

∆Tb = m.Kb
= 0,2 x 0,52
= 0,104

Tb = Tb° + ∆Tb
= 100 + 0,104
= 100,104°C

∆Tf = m.Kf
= 0,2 x 1,86
= 0,372°C

Tf = Tf° - ∆Tf
= 0 - 0,372
= -0,372°C