If a 0.570 𝑚 aqueous solution freezes at −2.00 ∘C, what is the van't Hoff factor, 𝑖, of the solute?

Answer 1

∆T = imK

∆T = change in freezing point = 2.00ºC

i = van't Hoff factor = ?

m = molality = 0.570 mol/kg

K = freezing constant for water = 1.86ºC/m

Solving for i ...

i = ∆T / (m)(K)

i = 2.00 / (0.570)(1.86)

i = 1.89