A solution of is observed to be exactly isotonic with a solution of glucose at the same temperature. — 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 $0.010\text{ M}$ solution of glucose at the same temperature. Calculate the apparent percentage degree of dissociation ($\% \alpha$) of the $Na_2SO_4$.
Answer: C
💡 Solution & Explanation
Isotonic means $i_1 C_1 = i_2 C_2$. Therefore, $i \times 0.004 = 1 \times 0.010 \implies i = 2.5$. For $Na_2SO_4$, $n=3$. Using $i = 1 + \alpha(n-1)$, we get $2.5 = 1 + 2\alpha \implies 1.5 = 2\alpha \implies \alpha = 0.75$, or $75\%$.
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