Terdapat 2 liter larutan glukosa 3M pada suhu 25C. Jika tekanan uap air pada suhu tersebut 23,76 mmHg Mr glukisa = 180, Mr air = 18 dan massa jenis larutan 1,2

gram glukosa = M . Mr . V
= 3 . 180 . 2
= 1080 gr
gram larutan = 1,2 gr/ml . 2000 ml
= 2400 gr
gram air = gr larutan - gr glukosa
= 2400 - 1080
= 1320 gr

P = Po . Xpelarut
= 23,76 . (1320/18)/(1320/18 + 1080/180)
= 23,76 . 0,92
= 21,96 mmHg

ΔP = Po - P
= 23,76 - 21,96
= 1,8 mmHg