Question

A molecule M associates in a given solvent according to the equation $M \Leftrightarrow \left( M \right)n$. For a certain concentration of M, the van't Hoff factor was found to be 0.9 and the fraction of associated molecules was 0.2. The value of n is:
A. 3
B. 5
C. 2
D. 4

Answer 1

The van 't Hoff factor is basically the ratio of the actual concentration of the particles produced on dissolving the substance and the concentration of a substance which is calculated from its mass.

Complete answer:
Van't Hoff introduced a factor i.e. 'i' is known as Van't Hoff's factor mainly to express the extent of association and dissociation of solutes in a solution. Degrees of association can be defined as the fraction of the total number of molecules that associate together or in other terms that combine together resulting into the formation of a bigger molecule.
Van’t Hoff’s factor is represented by iota. The formula for Van’t Hoff’s factor is generally expressed as follows in case of association:
$i = 1 - \alpha \left( {1 - \dfrac{1}{n}} \right)$ where i = Van’t Hoff’s factor, $\alpha$= fraction of associated molecules, n = number of molecules
In the question we are given:
i = 0.9
$\alpha$= 0.2
n = ?
Now, substituting the values in the formula, we will calculate the value of n:

$$0.9 = 1 - 0.2\left( {1 - \dfrac{1}{n}} \right) \\\ n = 2$$

**Hence, the correct answer is Option C.

Note:**
For normal liquids, the value of Van’t Hoff’s factor is 1, but it varies in case of association and dissociation of solutions. Always remember that for association, its value is usually less than 1 whereas for dissociation, its value is more than 1.