A molecule X associates in a given solvent as per the following equation:

X ⇌(X)_n

For a given concentration of X, the van’t Hoff factor was found to be 0.80 and the fraction of associated molecules was 0.3. The correct value of ‘n’ is:

The correct option is 3.

Understanding Association:
When molecules associate in a solvent, multiple individual molecules cluster together to form a larger aggregate. The association of molecule X can be represented by the following chemical equation:

$$n\mathrm{X}\rightleftharpoons {\left(\mathrm{X}\right)}_n$$

Here, n molecules of X associate to form one molecule of the polymer (X)_n.

Setting up the equilibrium:
Let the initial number of moles of X be 1, and the fraction of molecules that associate (degree of association) be α.

Initially:
Moles of X = 1
Moles of (X)_n = 0

At equilibrium:
Moles of unassociated X = 1 - α
Moles of associated polymer (X)_n = α / n

The total number of moles of particles at equilibrium is given by the sum of the moles of unassociated reactant and the associated product:

$$\text{Total moles at equilibrium}=\left(1-\alpha \right)+\frac{\alpha }{n}$$

Simplifying the equation:

$$\text{Total moles at equilibrium}=1-\alpha \left(1-\frac{1}{n}\right)$$

Van’t Hoff Factor (i):
The van’t Hoff factor (i) is defined as the ratio of the actual number of moles of particles at equilibrium to the initial number of moles of particles:

$$i=\frac{\text{Total moles at equilibrium}}{\text{Initial moles}}$$

$$i=1-\alpha \left(1-\frac{1}{n}\right)$$

Substituting the given values:
From the problem description, we are given:
- The van’t Hoff factor, i = 0.80
- The fraction of associated molecules (degree of association), α = 0.3

Substituting these values into the formula:

$$0.80=1-0.3\left(1-\frac{1}{n}\right)$$

Rearranging the equation to solve for n:

$$0.3\left(1-\frac{1}{n}\right)=1-0.80$$

$$0.3\left(1-\frac{1}{n}\right)=0.20$$

Divide both sides by 0.3:

$$1-\frac{1}{n}=\frac{0.20}{0.30}$$

$$1-\frac{1}{n}=\frac{2}{3}$$

Solve for 1/n:

$$\frac{1}{n}=1-\frac{2}{3}$$

$$\frac{1}{n}=\frac{1}{3}$$

Therefore, we get:

$$n=3$$

Thus, the correct value of ‘n’ is 3.