What is the freezing point of a solution that contains 0.550 moles of #NaI# in 615 g of water?

#K_f# = 1.86 C/m
molar mass of water = 18 g

1 Answer
Apr 11, 2017

Answer:

#-3.32^oC#

Explanation:

Freezing point depression is a function of the moles of solute in the moles of solvent. It is a “colligative property” based on particles in solution, not just compound molarity. First, we ‘normalize’ the given values to a standard liter of solution, using the density of water as #1g/(cm^3)#.

0.550/0.615L = 0.894 molar solution.

HOWEVER, in the case of NaI we have a compound that will dissociate completely into TWO moles of particles, doubling the “molar quantity” in solution.

Applying the freezing point depression constant for this compound we have:
#2 * 0.894 mol * 1.86( 'C)/(mol) = 3.32^oC# depression of the freezing point, or a change from #0’C# for water to #-3.32^oC#.