For solutes which do not undergo any association or dissociation in a solute, van't Hoff factor (i) will be

Options

A. less than 1
B. more than 1
C. equal to 1
D. zero

Answer

To solve the question regarding the van't Hoff factor (i) for solutes that do not undergo any association or dissociation in a solvent, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Van't Hoff Factor (i)**:
- The van't Hoff factor (i) is defined as the ratio of the number of particles of solute after dissociation or association to the number of particles of solute before dissociation or association.
- Mathematically, it can be expressed as:
\[ i = \frac{n_f}{n_i} \]
where \( n_f \) is the number of particles after dissociation/association, and \( n_i \) is the number of particles before dissociation/association.

2. **Identifying Non-Electrolytic Solutes**:
- Non-electrolytic solutes are those that do not dissociate into ions or associate into larger particles when dissolved in a solvent. Examples include sugar and urea.
- For these solutes, the number of particles remains the same before and after dissolving.

3. **Setting Up the Scenario**:
- Let’s assume we have a non-electrolytic solute, represented as AB. When this solute is dissolved in a solvent, it remains as AB without breaking into ions or forming aggregates.

4. **Calculating the Van't Hoff Factor**:
- Initially, if we have \( x \) moles of the solute AB, after dissolving, we still have \( x \) moles of AB.
- Therefore, we can substitute into the van't Hoff factor formula:
\[ i = \frac{n_f}{n_i} = \frac{x}{x} = 1 \]

5. **Conclusion**:
- The van't Hoff factor (i) for solutes that do not undergo any association or dissociation is equal to 1.

### Final Answer:
The van't Hoff factor (i) for solutes which do not undergo any association or dissociation in a solvent is **1**.