If in solvent, n simple molecules of solute combine to form an associated molecule `X` is degree of association the Van't Hoff's factor 'I' is equal to:

Options

A. `(1)/(1-nx)`
B. `(1-x-nx)/(1)`
C. `(1-x+(x)/(n))/(1)`
D. `((x)/(n)-1+x)/(1)`

Explanation

To find the Van't Hoff factor \( I \) in the case where \( n \) simple molecules of solute combine to form an associated molecule \( X \), we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Initial Condition**:
- Initially, we have 1 mole of solute and no associated molecules. Thus, at \( t = 0 \):
- Moles of solute = 1
- Moles of associated molecules = 0

2. **Defining the Change Over Time**:
- Let \( x \) be the number of moles of solute that have combined to form associated molecules at time \( t \).
- Therefore, the moles of solute remaining unassociated will be \( 1 - x \).
- The number of associated molecules formed will be \( \frac{x}{n} \) since \( n \) molecules combine to form 1 associated molecule.

3. **Calculating Moles After Association**:
- After some time \( t \):
- Moles of unassociated solute = \( 1 - x \)
- Moles of associated molecules = \( \frac{x}{n} \)
- Total moles after association = Moles of unassociated solute + Moles of associated molecules:
\[ \text{Total moles} = (1 - x) + \frac{x}{n} \]

4. **Finding the Van't Hoff Factor \( I \)**:
- The Van't Hoff factor \( I \) is defined as the ratio of the total number of moles after association to the initial number of moles:
\[ I = \frac{\text{Total moles after association}}{\text{Initial moles}} \]
- Substituting the values we found:
\[ I = \frac{(1 - x) + \frac{x}{n}}{1} \]
- Simplifying this gives:
\[ I = 1 - x + \frac{x}{n} \]

### Final Answer:
Thus, the Van't Hoff factor \( I \) is equal to:
\[ I = 1 - x + \frac{x}{n} \]