Van't Hoff Factor & its Significance

Van't Hoff factor (𝑖)

In 1880 Jacobus Henricus van’t Hoff introduced a factor (𝑖) , known as the van’t Hoff factor, to account for the extent to which a substance associates or dissociates in a solution.
The Van't Hoff factor tells about relationship between normal colligative properties and abnormal colligative properties.
It may be defined as the ratio of the number of solute particles in solution to the number of solute particles initially dissolved.

The Van’t Hoff factor is always a positive integer value; it can never be negative.
For a non-electrolyte solute, which doesn’t dissociate into ions in solution, the Van't Hoff factor is 1.
For electrolytes, which dissociate into ions, the Van't Hoff factor is greater than 1 and depends on the number of ions produced per formula unit of the solute.

Examples:
NaCl dissociates into two ions (Na⁺ and Cl⁻) in solution. So, its Van't Hoff factor is 2.
CaCl₂ dissociates into three ions (Ca⁺² and 2Cl⁻) in solution. So its Van't Hoff factor is 3.

The significant character of the Van't Hoff factor is it is used in the formulas for calculating colligative properties.
Relative lowering of vapour pressure of solvent, (P^o − P)/P^o = 𝑖 × (n₂/n₁)
Elevation of Boiling point, ΔT_b = 𝑖 × K_b × m
Depression of Freezing point, ΔT_f = 𝑖 × K_f × m
Osmotic pressure of solution, π = 𝑖 × C × R × T

Degree of dissociation(α) is related to the Van't Hoff factor(𝑖)as:
𝑖 = 1 + (n − 1)α
Degree of Association(α) is related to the Van't Hoff factor(𝑖)as:
𝑖 = (1 − α) + (α/n)
If α = 100% or 1 or not specified then-
𝑖 = 1/n

Which one has same Van't Hoff factor as that of Hg₂Cl₂

A. NaCl
B. Na₂SO₄
C. Al(NO₃)₃
D. Al₂(SO₄)₃

Answer

Option B is correct answer.

The Van't Hoff factor of Hg₂Cl₂ in its aqueous solution is 2.6 because salts like mercury chloride does not completely ionize; so, the number of ions dissociated may not equal the van't Hoff factor.
Hg₂Cl₂ is 80% dissociated so the degree of dissociation of α = 0.8
Hg₂Cl₂ ⇌ Hg₂⁺² + 2Cl⁻
n = 3
0.8 = (i − 1)/(3 − 1)
or, i = 2.6
2.6 is near to the i value of Na₂SO₄