Van't Hoff factor \((i)\) is given by

\(\frac{\text{Total number of moles of particles after dissociation}}{\text{Total number of moles of particles before dissociation}}\)

The correct answer is option 3. \(\frac{\text{Total number of moles of particles after dissociation}}{\text{Total number of moles of particles before dissociation}}\).

Let us break down the concept of the Van't Hoff factor \((i)\):

Colligative properties depend on the number of solute particles in a solution, not on their identity. These properties include:

• Elevation in Boiling point
• Depression in Freezing point
• Osmotic pressure
• Lowering of Vapor pressure

The Van't Hoff factor \((i)\) is a correction factor that accounts for the actual number of particles in solution compared to what was initially expected based on the solute's formula. It is particularly important when solutes undergo dissociation (breaking into more particles) or association (combining into fewer particles) in solution.

The Van't Hoff factor is defined as:

\(i = \frac{\text{Total number of moles of particles after dissociation/association}}{\text{Total number of moles of particles before dissociation/association}}\)

Before dissociation/association: This refers to the number of solute particles you would expect based on the chemical formula. For example, 1 mole of \( \text{NaCl} \) before dissociation would be 1 mole of particles.

After dissociation/association: This refers to the actual number of particles in solution after the solute has dissociated or associated. For example, 1 mole of \( \text{NaCl} \) dissociates into 1 mole of \( \text{Na}^+ \) and 1 mole of \( \text{Cl}^- \), resulting in 2 moles of particles.

Dissociation Example: Consider \( \text{NaCl} \):

\(\text{NaCl} \rightarrow \text{Na}^+ + \text{Cl}^-\)

Before dissociation: 1 mole of \( \text{NaCl} \) corresponds to 1 mole of particles.

After dissociation: 1 mole of \( \text{NaCl} \) produces 2 moles of particles (\(\text{Na}^+\) and \(\text{Cl}^-\)).

Van't Hoff factor \(i = \frac{2 \text{ moles}}{1 \text{ mole}} = 2\).

Association Example: Consider a dimerization reaction where 2 molecules of a solute combine to form one molecule in solution:

\(2A \rightarrow A_2\)

Before association: 2 moles of \(A\).

After association: 1 mole of \(A_2\).

Van't Hoff factor \(i = \frac{1 \text{ mole}}{2 \text{ moles}} = 0.5\).

The Van't Hoff factor directly affects the magnitude of colligative properties. For example, if \(i > 1\), the effect on freezing point depression or boiling point elevation is greater than expected because there are more particles in solution. If \(i < 1\), the effect is less than expected due to fewer particles in solution (association).

The Van't Hoff factor is defined as: \(i = \frac{\text{Total number of moles of particles after dissociation/association}}{\text{Total number of moles of particles before dissociation/association}}\)

So, the correct choice is option 3: \(\frac{\text{Total number of moles of particles after dissociation}}{\text{Total number of moles of particles before dissociation}}\).