Question

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

Answer 1

`-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`.