Question:medium

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

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Normality = Molarity $\times$ n-factor. For acids, n-factor is the number of replaceable hydrogens.
  • 0.266 N
  • 0.133 N
  • 0.399 N
  • 0.430 N
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The Correct Option is B

Solution and Explanation

Step 1: Understanding the Concept:
Normality ($N$) is defined as the number of gram equivalents of solute per liter of solution. For oxalic acid ($H_2C_2O_4 \cdot 2H_2O$), it is a dibasic acid, meaning it can donate two protons ($H^+$).
Step 2: Key Formula or Approach:
1. Equivalent weight = $\frac{\text{Molecular weight}}{\text{Basicity (n-factor)}}$.
2. Normality ($N$) = $\frac{\text{Mass (g)}}{\text{Equivalent weight}} \times \frac{1000}{\text{Volume (mL)}}$.
Step 3: Detailed Explanation:
Oxalic acid dihydrate has a molecular weight of 126 and an n-factor of 2. \[ \text{Equivalent weight} = \frac{126}{2} = 63 \] Now, calculate Normality: \[ N = \frac{12.6}{63} \times \frac{1000}{1500} \] \[ N = 0.2 \times \frac{2}{3} = \frac{0.4}{3} \approx 0.1333 \text{ N} \]
Step 4: Final Answer:
The normality of the solution is 0.133 N.
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