Sebanyak 18g glukosa c6 H12O6 dilarutkan dalam air 200g jika kb=0,52 dan kf 1,86.ar c = 12 o=16 h=1 tentukan titik didih larutan

wt = 18 gram
wp = 200 gram
Kb = 0,52
Kf = 1,86
Mr = 180
Tb = .......................?

ΔTb = wt/Mr x 1000/wp x Kb
= 18/180 x 1000/200 x 0,52
= 0,1 x 5 x 0,52
= 0,26°C

Tb = Tb° + ΔTb
= 100 + 0,26
= 100,26°C