Question:
At T(K), 0.004 M $Na_2SO_4$ solution is isotonic with 0.01 M glucose solution. The degree of dissociation of $Na_2SO_4$ is:
At T(K), 0.004 M $Na_2SO_4$ solution is isotonic with 0.01 M glucose solution. The degree of dissociation of $Na_2SO_4$ is:
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van't Hoff factor $i = 1 + (n-1)\alpha$ for dissociation.
Updated On: Jun 6, 2026
- 80 %
- 50 %
- 25 %
- 75 %
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The Correct Option is D
Solution and Explanation
Step 1: Concept
Isotonic solutions have equal osmotic pressure: $i_1 C_1 = i_2 C_2$.
Step 2: Meaning
Glucose ($i=1$), $Na_2SO_4 \rightarrow 2Na^+ + SO_4^{2-}$ ($i = 1 + (n-1)\alpha$).
Step 3: Analysis
$1 \times 0.01 = [1 + (3-1)\alpha] \times 0.004$. $0.01 = 0.004 \times (1 + 2\alpha) \rightarrow 2.5 = 1 + 2\alpha \rightarrow 1.5 = 2\alpha \rightarrow \alpha = 0.75$. $\alpha = 75\%$.
Step 4: Conclusion
Degree of dissociation is 75
Final Answer: (D)
Isotonic solutions have equal osmotic pressure: $i_1 C_1 = i_2 C_2$.
Step 2: Meaning
Glucose ($i=1$), $Na_2SO_4 \rightarrow 2Na^+ + SO_4^{2-}$ ($i = 1 + (n-1)\alpha$).
Step 3: Analysis
$1 \times 0.01 = [1 + (3-1)\alpha] \times 0.004$. $0.01 = 0.004 \times (1 + 2\alpha) \rightarrow 2.5 = 1 + 2\alpha \rightarrow 1.5 = 2\alpha \rightarrow \alpha = 0.75$. $\alpha = 75\%$.
Step 4: Conclusion
Degree of dissociation is 75
Final Answer: (D)
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