If a 0.500 𝑚 aqueous solution freezes at −2.20 ∘C, what is the van't Hoff factor, 𝑖, of the solute?

Answer 1

∆T = imK

∆T = change in freezing point = 2.20º

i = van't Hoff factor = ?

m = molality = 0.500 m

K = freezing constant for water = 1.86 m/º

Solving for i, we have...

i = ∆T / (m)(K)

i = 2.20 / (0.500)(1.86)

i = 2.37