Hitunglah titik didih dan titik beku larutan yang mengandung 50 gram glukosa Mr = 180dalam 500 gram air. Kb air=0,52 0C/m kf air =

Penjelasan:

diketahui

gr = 50

Mr = 180

p = 500

kb = 0,52

kf = 1,86

i = 1 (larutan non elektrolit)

∆Tb = gr/Mr × 1000/p × kb× i

∆Tb = 50/180 × 1000/500 × 0,52 × 1

∆Tb = 0,288

Tb = 100+∆Tb = 100+0,288

Tb = 100,288°C ——–> Titik didih

∆Tf = gr/Mr × 1000/p × kf × i

∆Tf = 50/180 × 1000/500 × 1,86 × 1

∆Tf = 1,033

Tf = 0-∆Tf = 0-1,033

Tf= -1,033°C —-> titik beku