This question concerns freezing point depression.

Question

Aqueous solutions with a molality of 0.020 m are prepared from Na3PO4, KCl, C6H12O6, and CaCl2. Determine the expected freezing point of each solution.

Answer 1

The freezing point depression is given by...

∆T = imK

∆T = change in freezing point = ?

i = van't Hoff factor (see below for each compound)

m = molality = moles solute / kg solvent = 0.020

K = freezing point depression constant for water = 1.86º/m

For Na₃PO₄

Na₃PO₄ ==> 3Na⁺ + PO₄^3- (i = 4)

∆T = (4)(0.020)(1.86) = 0.149º

Freezing point of solution = -0.15ºC

For KCl

KCl ==> K⁺ + Cl^- (i = 2)

∆T = (2)(0.020)(1.86) = 0.0744

Freezing point of solution = -0.074º

Apply same calculations for the following:

For C₆H₁₂O₆

C₆H₁₂O₆ ==> C₆H₁₂O₆ (i = 1)

For CaCl₂

CaCl₂ ==> Ca²⁺ + 2Cl^- (i = 3)

Answer 2

Hi Ivy E.,

Here you are spamming Quora again. Please check your textbook or just reference Google: molality is moles of solvent per kg of solvent. The composition of the solute is irrelevant, if the molality is given to you. There are some things (ionic strength, and things dependent on that, such as activity coefficients) for which number of ions per nominal formula unit is important, but colligative properties are not one of such things.

-- Cheers, --Mr. d.