Larutan elektrolit 0,320 gram CuSO4 dilarutkan ke dalam 2 liter air. Jika massa jenis air = 1 g/mL, kf air = 1,86 °C/m. maka hitunglah Titik beku larutan terseb

Diketahui :
- massa = 0,32 gram
- v = 2 liter = 2000 ml
- kf = 1,86 °C/m
- Mr CuSO4 = 160 - CuSO4 = Cu²+ + SO4²- n = 2 ➡ i = {1+(n-1)a} = {1+(2-1)1) = 2
Ditanya : Tf larutan?
Jawab :
m = massa/Mr x 1000/v
= 0,32/160 x 1000/2000
= 0,001 molar

∆Tf = Kf x m x i
= 1,86 x 0,001 x 2
= 0,00372°C

∆Tf = Tf pelarut - Tf larutan
Tf larutan = Tf pelarut - ∆Tf
= 0 - 0,00372
= -0,00372°C
Semoga membantu;))