How do you calculate freezing point depression?

1 Answer
Dec 31, 2013

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.