Question

12.6 g of oxalic acid, $H_{2}C_{2}O_{4} \cdot 2H_{2}O$ (M.wt 126) is present in 1500 mL of solution. The normality of that solution is

• A. 0.266 N
• B. 0.133 N
• C. 0.399 N
• D. 0.430 N

Hint:

Normality = Molarity $\times$ n-factor. For acids, n-factor is the number of replaceable hydrogens.

Answer 1

The Correct Option is B

Step 1: Concept
Normality ($N$) is defined as the number of gram equivalents of solute per liter of solution.

Step 2: Meaning
For oxalic acid ($H_{2}C_{2}O_{4} \cdot 2H_{2}O$), the n-factor (basicity) is 2 because it provides two $H^{+}$ ions. Equivalent weight = Molecular weight n-factor = $126 / 2 = 63$.

Step 3: Analysis
Number of gram equivalents = Mass Equivalent weight = $12.6 / 63 = 0.2$. Volume of solution in liters = $1500 / 1000 = 1.5$ L. Normality = Equivalents Volume (L) = $0.2 / 1.5$.

Step 4: Conclusion
Normality = $0.1333...$ N, which matches Option (B).
Final Answer: (B)