A solution of is observed to be exactly isotonic with a aqueous solution of glucose at the same temp — Solutions and Colligative Properties Chemistry Question
Question
A $0.004\text{ M}$ solution of $Na_2SO_4$ is observed to be exactly isotonic with a $0.010\text{ M}$ aqueous solution of glucose at the same temperature. Assuming molarity is approximately equal to molality, what is the apparent percentage degree of dissociation ($\% \alpha$) of $Na_2SO_4$?
💡 Solution & Explanation
Isotonic solutions have identical osmotic pressures ($\pi_1 = \pi_2$), meaning their effective concentrations ($i \times C$) are equal. $i \times 0.004 = 1 \times 0.010 \implies i = 2.5$. For $Na_2SO_4$, $n=3$ ions ($2 Na^+$ and $SO_4^{2-}$). Using $i = 1 + \alpha(n-1) \implies 2.5 = 1 + \alpha(3-1) \implies 2.5 = 1 + 2\alpha \implies \alpha = 0.75$ or $75\%$.