Question

Arrange the following in order of their decreasing ionic strength: A. 0.1 m NaCl, B. 0.01 m $K_{2}SO_{4}$, C. 0.001 m $AlCl_{3}$, D. 0.01 m glucose.}

• A. B, C, A, D
• B. D, C, B, A
• C. A, B, C, D
• D. C, B, D, A

Hint:

For 1:1 electrolytes like NaCl, Ionic Strength ($I$) is exactly equal to the Molality ($m$).

Answer 1

The Correct Option is C

Step 1: Concept Ionic strength ($I$) is calculated as $I = \frac{1}{2} \sum m_{i} z_{i}^{2}$, where $m$ is molality and $z$ is the charge of the ion.

Step 2: Meaning Higher molality and higher charges on ions lead to higher ionic strength. Non-electrolytes have an ionic strength of zero.

Step 3: Analysis For A (NaCl): $I = \frac{1}{2}[0.1(1^{2}) + 0.1(1^{2})] = 0.1$. For B ($K_{2}SO_{4}$): $I = \frac{1}{2}[0.02(1^{2}) + 0.01(2^{2})] = 0.03$. For C ($AlCl_{3}$): $I = \frac{1}{2}[0.001(3^{2}) + 0.003(1^{2})] = 0.006$. For D (Glucose): Non-electrolyte, $I = 0$.

Step 4: Conclusion Comparing the values: $0.1 > 0.03 > 0.006 > 0$. The order is A $>$ B $>$ C $>$ D. Final Answer: (C)