Question

A molecule XY (mol. wt $180~g~mol^{-1}$) undergoes second order association in water. If the van't Hoff factor 'i' of the substance is 0.88; the degree of association of the substance is:

• A. 10%
• B. 28%
• C. 48%
• D. 24%

Hint:

For dimerization, the degree of association $\alpha$ is always $2(1-i)$.

Answer 1

The Correct Option is D

Step 1: Concept For association of $n$ molecules into one ($nXY \rightarrow (XY)_{n}$), the van't Hoff factor is $i = 1 - \alpha(1 - \frac{1}{n})$, where $\alpha$ is the degree of association.

Step 2: Meaning "Second order association" means $n = 2$ (dimerization).

Step 3: Analysis Given $i = 0.88$ and $n = 2$, we set up the equation: $0.88 = 1 - \alpha(1 - \frac{1}{2})$. This simplifies to $0.88 = 1 - \frac{\alpha}{2}$. Rearranging: $\frac{\alpha}{2} = 1 - 0.88 = 0.12$. Thus, $\alpha = 0.24$.

Step 4: Conclusion Converting to percentage: $0.24 \times 100 = 24%$. Final Answer: (D)