Assuming equal concentrations and complete dissociation, rank these aqueous solutions by their freezing points. li2so4 nh4i cocl3

CoCl₃ > Li₂SO₄ > NH₄I.

Explanation:

Adding solute to water causes depression of the boiling point.The elevation in boiling point (ΔTf) can be calculated using the relation:

ΔTf = i.Kf.m,

where, ΔTf is the depression in freezing point.

i is the van 't Hoff factor.

van 't Hoff factor is the ratio between the actual concentration of particles produced when the substance is dissolved and the concentration of a substance as calculated from its mass. For most non-electrolytes dissolved in water, the van 't Hoff factor is essentially 1.

Kf is the molal depression constant of water.

m is the molality of the solution.

(1) Li₂SO₄:

i for Li₂SO₄ = no. of particles produced when the substance is dissolved/no. of original particle = 3/1 = 3.

∴ ΔTb for (Li₂SO₄) = i.Kb.m = (3)(Kf)(m) = 3(Kf)(m).

(2) NH₄I:

i for NH₄I = no. of particles produced when the substance is dissolved/no. of original particle = 2/1 = 2.

∴ ΔTb for (NH₄I) = i.Kb.m = (2)(Kf)(m) = 2(Kf)(m).

(3) CoCl₃:

i for CoCl₃ = no. of particles produced when the substance is dissolved/no. of original particle = 4/1 = 4.

∴ ΔTb for (CoCl₃) = i.Kb.m = (4)(Kf)(m) = 4(Kf)(m).

So, the ranking of the freezing point from the highest to the lowest is:

CoCl₃ > Li₂SO₄ > NH₄I.

30litre of li2o is 1.0002mole approximately 1 mole

The answer is the first option just took the test