If a solution of 1.00g of unknown solute is added to 9.116 g of steric acid and the solution freezes at 63.6 oC, What is the molality of the solution

Question

( Normal FP steric acid = 68.88oC and Kf = 4.50 oC/m )

Answer 1

Colligative Properties:

Freezing point depression.

∆T = imK

∆T = change in freezing point = 68.88º - 63.6º = 5.28º

i = van't Hoff factor = 1 if the unknown doesn't ionize or dissociate. They don't mention the nature of the unknown so we'll have to assume the value of i to be 1. This is a poorly written quesiton.

m = molality = moles solute / kg solvent= ?

K = freezing constant = 4.50º/m

5.28 = (1)(m)(4.50)

m = 1.17 m (the fact that 1.00 g is dissolved in 9.116 g does not enter into the calculation, unless you wanted to determine the molar mass of the unknown)

[What a strange question, for sure!]