Association Equation Calculation

The association of molecule X is given by the equilibrium: $X \rightleftharpoons (X)_n$

The van't Hoff factor ($i$) relates the observed colligative property to the theoretical one, accounting for association or dissociation.

For association, the formula for the van't Hoff factor is:

$i = 1 - \alpha + \frac{\alpha}{n}$

Where:

• $i$ is the van't Hoff factor.
• $\alpha$ is the fraction of the solute that has associated.
• $n$ is the number of molecules that associate to form one larger molecule.

Applying Given Values

From the question, we have:

• Van't Hoff factor, $i = 0.80$.
• Fraction of associated molecules, $\alpha = 0.3$.

We need to find the value of $n$.

Solving for 'n'

Substitute the given values into the equation:

$0.80 = 1 - 0.3 + \frac{0.3}{n}$

Simplify the equation:

$0.80 = 0.7 + \frac{0.3}{n}$

Isolate the term containing $n$:

$\frac{0.3}{n} = 0.80 - 0.7$

$\frac{0.3}{n} = 0.10$

Solve for $n$:

$n = \frac{0.3}{0.10}$

$n = 3$

Thus, the correct value of $n$ is 3.