Sebanyak 18 gram glukosa C6H12O6 (Mr =180) dilarutkan kedalam 500 gram air jika kf = 1.8°c\ mol titik beku larutan tersebut adalah...

a = 18 gram  →  (massa zat terlarut)
Mr = 180  → (Mr zat terlarut)
p = 500 gram → (massa pelarut)
Kf  air = 1,86° C/molal
ΔTf = ?


ΔTf = 1,86 x 18/180 x 1000/500
= 1,86 x 0,1 x 2 = 0,372°C