A specific organic molecule heavily associates in a non-polar solvent according to the equilibrium e — Solutions and Colligative Properties Chemistry Question
Question
A specific organic molecule $M$ heavily associates in a non-polar solvent according to the equilibrium equation $n M \rightleftharpoons (M)_n$. For a specific concentrated solution of $M$, the van't Hoff factor ($i$) was experimentally determined to be $0.90$. If the fraction of the molecules undergoing association ($\alpha$) is precisely $0.20$, what is the exact numerical value of $n$ (the degree of polymerization)?
💡 Solution & Explanation
For the association process $n M \rightleftharpoons (M)_n$, the van't Hoff factor is theoretically given by $i = 1 + \alpha(\frac{1}{n} - 1)$. We are given $i = 0.90$ and $\alpha = 0.20$. Substituting these values: $0.90 = 1 + 0.20(\frac{1}{n} - 1) \implies -0.10 = \frac{0.20}{n} - 0.20 \implies 0.10 = \frac{0.20}{n}$. Solving for $n$ gives $n = \frac{0.20}{0.10} = 2$, indicating a dimerization process.