Sebanyak 45 g glukosa (Mr=180) dilarutkan dalam 400 g air. Jika diketahui Kf air 1,86 °C/m,tentukan titik beku larutan glukosa

Mr = 180
p = 400 gram
g = 45 gram
kf = 1,86

∆Tf = m . kf
∆Tf = (45/180) × (1000/400) × 1,86
∆Tf = ¼ × 5/2 × 1,86
∆Tf = 9,3/8
∆Tf = 1,16

Tf = 0 – ∆Tf
Tf = -1,16°C