# 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

Apr 11, 2017

$- {3.32}^{o} C$

#### 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 $1 \frac{g}{c {m}^{3}}$.

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:
$2 \cdot 0.894 m o l \cdot 1.86 \frac{' C}{m o l} = {3.32}^{o} C$ depression of the freezing point, or a change from 0’C for water to $- {3.32}^{o} C$.