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QUESTION

# How do you calculate 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

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.