Question:
\(12.6\ \text{g}\) of oxalic acid, \(H_2C_2O_4\cdot 2H_2O\) \((\text{M.wt. }126)\), is present in \(1500\ \text{mL}\) of solution. The normality of that solution is
\(12.6\ \text{g}\) of oxalic acid, \(H_2C_2O_4\cdot 2H_2O\) \((\text{M.wt. }126)\), is present in \(1500\ \text{mL}\) of solution. The normality of that solution is
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For acids, equivalent weight \(=\frac{\text{molecular weight}}{\text{basicity}}\). Oxalic acid has basicity \(2\).
Updated On: May 7, 2026
- \(0.266\ N\)
- \(0.133\ N\)
- \(0.399\ N\)
- \(0.430\ N\)
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The Correct Option is B
Solution and Explanation
Concept:
Normality is:
\[
N=\frac{\text{gram equivalents}}{\text{volume in litres}}
\]
For oxalic acid, \(n\)-factor is \(2\).
Step 1: Given molecular weight: \[ 126 \] For oxalic acid: \[ n=2 \] Equivalent weight: \[ \frac{126}{2}=63 \]
Step 2: Given mass: \[ 12.6\ \text{g} \] Gram equivalents: \[ \frac{12.6}{63}=0.2 \]
Step 3: Volume: \[ 1500\ \text{mL}=1.5\ \text{L} \]
Step 4: Normality: \[ N=\frac{0.2}{1.5} \] \[ N=0.133 \] Therefore, \[ \boxed{0.133\ N} \]
Step 1: Given molecular weight: \[ 126 \] For oxalic acid: \[ n=2 \] Equivalent weight: \[ \frac{126}{2}=63 \]
Step 2: Given mass: \[ 12.6\ \text{g} \] Gram equivalents: \[ \frac{12.6}{63}=0.2 \]
Step 3: Volume: \[ 1500\ \text{mL}=1.5\ \text{L} \]
Step 4: Normality: \[ N=\frac{0.2}{1.5} \] \[ N=0.133 \] Therefore, \[ \boxed{0.133\ N} \]
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