dalam 250 gram air dilarutkan 1,9 gram MgCl2 ternyata larutan membeku pada suhu -0,372 derajat celsius tetapan titik beku molal air = 1,86derajat celcius/m. der

MgCl₂ → Mg²⁺ + 2Cl⁻ (n = 3)
wp = 250 gram
wt = 1,9 gram
Tf = -0,372°C
Kf = 1,86
Mr = 95
α = ..................?

ΔTf = Tf° - Tf
= 0 - (-0,372)
= 0,372°C

ΔTf = wt/Mr x 1000/wp x Kf x i
0,372 = 1,9/71 x 1000/250 x 1,86 x i
0,372 = 0,027 x 4 x 1,86 x i
0,372 = 0,201 x i
i = 0,372/0,201
i = 1,85

i = 1+(n-1)α
1,85 = 1+(3-1)α
1,85 = 1+2α
2α = 1,85 - 1
α = 0,85/2
α = 0,425
α = 0,43