Van't Hoff factor — why does it differ for electrolytes and non-electrolytes

Question

What is the Van’t Hoff factor $i$ ? Explain why $i > 1$ for electrolytes and $i < 1$ for substances that associate in solution. Calculate $i$ for 0.1 M NaCl that is 90% dissociated.

(JEE Main 2023, similar pattern)

Solution — Step by Step

The Van’t Hoff factor $i$ is the ratio of the observed colligative property to the calculated colligative property (assuming no dissociation or association):

i = \frac{\text{Observed colligative property}}{\text{Calculated colligative property}}

It corrects colligative property formulas for electrolytes:

\Delta T_b = i \cdot K_b \cdot m, \quad \pi = iCRT

Electrolytes dissociate in solution, producing **more particles** than the formula suggests. NaCl → Na$^+$ + Cl$^-$ gives 2 particles from 1 formula unit.

Colligative properties depend on the number of solute particles, not their nature. More particles → larger colligative effect → $i > 1$ .

For complete dissociation of NaCl: $i = 2$ . For $\text{CaCl}_2$ : $i = 3$ . For $\text{AlCl}_3$ : $i = 4$ .

Some solutes (like acetic acid in benzene) associate — two or more molecules combine to form a single unit. This reduces the effective number of particles.

Fewer particles → smaller colligative effect → $i < 1$ .

NaCl dissociates as: NaCl → Na $^+$ + Cl $^-$

If $\alpha = 0.9$ (90% dissociation), and NaCl produces $n = 2$ ions:

i = 1 + \alpha(n - 1) = 1 + 0.9(2 - 1) = 1 + 0.9 = \mathbf{1.9}

If it were 100% dissociated, $i = 2$ . At 90%, it is 1.9.

Why This Works

Colligative properties (boiling point elevation, freezing point depression, osmotic pressure, vapor pressure lowering) depend solely on particle count. The Van’t Hoff factor adjusts for the actual number of particles in solution.

The formula $i = 1 + \alpha(n-1)$ is derived by counting: from 1 mole of solute, $\alpha$ fraction dissociates into $n$ ions each, and $(1-\alpha)$ fraction remains undissociated. Total particles = $(1-\alpha) + n\alpha = 1 + \alpha(n-1)$ .

Alternative Method — Direct Particle Counting

For 1 mol of NaCl with 90% dissociation:

Undissociated NaCl: $1 - 0.9 = 0.1$ mol
Na $^+$ ions: $0.9$ mol
Cl $^-$ ions: $0.9$ mol
Total particles: $0.1 + 0.9 + 0.9 = 1.9$ mol

So $i = 1.9/1 = 1.9$ .

For NEET, memorise the $i$ values for common salts assuming complete dissociation: NaCl, KCl → 2; $\text{MgCl}_2$ , $\text{Na}_2\text{SO}_4$ → 3; $\text{AlCl}_3$ , $\text{K}_3[\text{Fe(CN)}_6]$ → 4. Then multiply by the degree of dissociation if given.

Common Mistake

Students often confuse the number of ions $n$ with the Van’t Hoff factor $i$ . They are not the same unless dissociation is 100%. $n$ is the theoretical maximum number of particles per formula unit; $i$ is the actual effective number accounting for incomplete dissociation. Always use $i = 1 + \alpha(n-1)$ when $\alpha < 1$ .

Question

Solution — Step by Step

Why This Works

Alternative Method — Direct Particle Counting

Common Mistake

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