Question

How do you calculate freezing point depression?

Answer 1

You use the formula for freezing point depression.

Explanation:

EXAMPLE

What is the freezing point depression caused by adding 31.65 g of sodium chloride to 220.0 g of water. `K_"f"` for water is `"1.86 °C·kg·mol"^"-1"`.

Solution

The formula for freezing point depression expression is

`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_"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

`"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"`

`"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 = "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 `"NaCl"` completely dissociate into ions.

`"NaCl" → "Na"^+ + "Cl"^"-"`

One mole of solid `"NaCl"` gives two moles of dissolved particles: 1 mol of `"Na"^+` ions and 1 mol of `"Cl"^"-"` ions. Thus, for `"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.