Consider 0.01 m aqueous solutions of each of the following. a) NaI; b) CaCl2; c) K3PO4; and d) C6H12O6 (glucose) Arrange the solutions in order of freezing point from lowest to highest. Assume that each compound behaves ideally.

The solutions are ordered by this way (from lowest to highest freezing point):  K₃PO₄ < CaCl₂ < NaI < glucose

Option d, b, a and c

Explanation:

Colligative property: Freezing point depression

The formula is: ΔT = Kf . m . i

ΔT = Freezing T° of pure solvent - Freezing T° of solution

We need to determine the i, which is the numbers of ions dissolved. It is also called the Van't Hoff factor.

Option d, which is glucose is non electrolyte so the i = 1

a. NaI →  Na⁺  +  I⁻        i =2

b. CaCl₂ →  Ca²⁺  +  2Cl⁻      i =3

c. K₃PO₄ → 3K⁺ + PO₄⁻³     i=4

Potassium phosphate will have the lowest freezing point, then we have the calcium chloride, the sodium iodide and at the end, glucose.

consider the reaction below.

 no rx

at 500 k, the reaction is at equilibrium with the following concentrations.

[pci5]= 0.0095 m

[pci3] = 0.020

[ci2] = 0.020 m

what is the equilibrium constant for the given reaction?

0.042

0.42

2.4

24