Chemistry freezing point

Question

The normal freezing point of acetic acid is -45.3 °C. What is the new freezing point of acetic acid when 12.543 g of potassium sulfate (MM = 174.26 g/mol) is dissolved in 500.0 mL of acetic acid? What is the molality of the potassium sulfate solution? The freezing point constant of acetic acid is 2.20 °C/m, and the

density of acetic acid is 0.9970 g/mL

Answer 1

So, for this type of problem you'll want to use ...

∆T = imK

∆T = change in freezing point = ?

i = van't Hoff factor = 3 for potassium sulfate (K₂SO₄ => 2K⁺ + 1SO₄^2-, i.e. 3 particles)

m = molality = moles K₂SO₄ / kg acetic acid (see below for calculation of m)

K = freezing point constant for acetic acid = 2.20 ºC

Calculation of m (moles K₂SO₄ / kg acetic acid)

12.543 g xK₂SO₄ x 1 mol K₂SO₄ / 174.26 g = 0.017198 moles K₂SO₄

500.0 ml acetic acid x 0.9970 g / ml = 498.50 g acetic acid = 0.49850 kg acetic acid

molarity = 0.017198 mols / 0.49850 mols = 0.0344995 m

Calculate change in freezing point:

∆T = (3)(0.0344995 m)(2.20 ºC/m)

∆T = 0.228 (3 sig. figs. based on 2.20 ºC/m)

Calculate new freezing point of the solution:

-45.3º + -0.228º = -45.5ºC