# How do you calculate freezing point depression?

Dec 31, 2013

#### Explanation:

EXAMPLE

What is the freezing point depression caused by adding 31.65 g of sodium chloride to 220.0 g of water. ${K}_{\text{f}}$ for water is $\text{1.86 °C·kg·mol"^"-1}$.

Solution

color(blue)(|bar(ul(color(white)(a/a) ΔT_"f" = iK_"f"m color(white)(a/a)|)))" "

where

ΔT_"f" is the freezing point depression,

$i$ is the van’t Hoff factor,

${K}_{\text{f}}$ is the molal freezing point depression constant for the solvent, and

$m$ is the molality of the solution.

Step 1: Calculate the molality of the NaCl

$\text{moles of NaCl" = 31.65 color(red)(cancel(color(black)("g NaCl"))) × (1"mol NaCl")/(58.44 color(red)(cancel(color(black)("g NaCl")))) = "0.5416 mol NaCl}$

$\text{mass of water" = 220.0 color(red)(cancel(color(black)("g H"_2"O"))) × (1"kg H"_2"O")/(1000 color(red)(cancel(color(black)("g H"_2"O")))) = "0.220 kg H"_2"O}$

$m = \text{moles of NaCl"/"kilograms of water" = "0.5416 mol"/"0.220 kg" = "2.46 mol/kg}$

Step 2: Determine the van't Hoff factor

The van't Hoff factor, $i$, is the number of moles of particles obtained when 1 mol of a solute dissolves.

Nonelectrolytes such as sugar do not dissociate in water. One mole of solid sugar gives one mole of dissolved sugar molecules.

For nonelectrolytes, $i = 1$.

Electrolytes such as $\text{NaCl}$ completely dissociate into ions.

$\text{NaCl" → "Na"^+ + "Cl"^"-}$

One mole of solid $\text{NaCl}$ gives two moles of dissolved particles: 1 mol of ${\text{Na}}^{+}$ ions and 1 mol of $\text{Cl"^"-}$ ions. Thus, for $\text{NaCl} , i = 2$.

Step 3: Calculate ΔT_"f"

ΔT_"f" = iK_"f"m = 2 × "1.86 °C·"color(red)(cancel(color(black)("kg·mol"^"-1"))) × 2.46 color(red)(cancel(color(black)("mol·kg"^"-1"))) = "9.16 °C"

The freezing point depression is 9.16 °C.

Here's a video on how to calculate freezing point depression.