Answer: CoBr3 < K2SO4 < NH4 Cl

Justification:

1) The depression of the freezing point of a solution is a colligative property, which means that it depends on the number of particles of solute dissolved.

2) The formula for the depression of freezing point is:

ΔTf = i * Kf * m

Where i is the van't Hoof factor which accounts for the dissociation of the solute.

Kf is the freezing molal constant and only depends on the solvent

m is the molality (molal concentration).

3) Since, you are assuming equal concentrations and complete dissociation of the given solutes, the solute with more ions in the molecular formula will result in the solution with higher depression of the freezing point (lower freezing point).

4) These are the dissociations of the given solutes:

a) NH4 Cl (s) --> NH4(+)(aq) + Cl(-) (aq) => 1 mol --> 2 moles

b) Co Br3 (s) --> Co(3+) (aq) + 3Br(-)(aq) => 1 mol --> 4 moles

c) K2SO4 (s) --> 2K(+) (aq) + SO4 (2-) (aq) => 1 mol --> 3 moles

5) So, the rank of solutions by their freezing points is:

CoBr3 < K2SO4 < NH4 Cl

4 0

3 years ago

Read 2 more answers

G determine the concentration of an hbr solution if a 45.00 ml aliquot of the solution yields 0.6485 g agbr when added to a solu

Sunny_sXe [5.5K]

The molecular weight of silver bromide (AgBr) is 187.77 g/mole. The presence of the ions in solution can be shown as- AgBr (insoluble) ⇄ $Ag^{+}$ + $Br^{-1}$ .

45.00 mL of the aliquot contains 0.6485 g of AgBr. Thus 1000 mL of the aliquot contains $\frac{0.6485}{45}$ ×1000 = 14.411 gm-mole. Thus the solubility product $K_{sp}$ of AgBr = [ $Ag^{+}$ ]× $Br^{-}$ .