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# How do you calculate freezing point depression?

You use the formula for freezing point depression.

**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 , and

##m## is the 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 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.