Pyrolusite is the main ore of manganese in which it is present as Its Mn content is determined by reducing it under acidic condition to `Mn^(2+)` with the help of oxalate `(C_(2)O_(4)^(2-))` ion which in turn gets oxidized to `CO_(2)` The analytical determination is carried out by adding a known excess volume of `(C_(2)O_(4)^(2-))` solution to a suspension of the pyrolusite and digesting the mixture on a hot water bath until all the `MnO_(2)` has been reduced. The excess unreacted oxalate solution is then titrated with standardized `KMnO_(4)` solution. Thereby Mn content of ore can be calculated. `KMnO_(4)`solution is also standardized under acidic condition against oxalate ion wherein `MnO_(4)^(-)` ion is reduced to `Mn^(2+)` and `(C_(2)O_(4)^(2-))` ion is oxidized to `CO_(2)`
Q If a student prepared a standard solution of `Na_(2)C_(2)O_(4)` by dissolving 3.2 g of dry anhydrous salt into distilled water and making the solution upto 500 mL. The Normality of oxalate solution is

Options

A. 0.085
B. 0.096
C. 0.054
D. 0.032

Explanation

To find the normality of the oxalate solution prepared by dissolving 3.2 g of Na₂C₂O₄ in distilled water to make a 500 mL solution, we can follow these steps:

### Step 1: Calculate the Molar Mass of Na₂C₂O₄
The molar mass of Na₂C₂O₄ (sodium oxalate) can be calculated as follows:
- Sodium (Na): 22.99 g/mol × 2 = 45.98 g/mol
- Carbon (C): 12.01 g/mol × 2 = 24.02 g/mol
- Oxygen (O): 16.00 g/mol × 4 = 64.00 g/mol

Adding these together:
\[ \text{Molar mass of Na}_2\text{C}_2\text{O}_4 = 45.98 + 24.02 + 64.00 = 134.00 \, \text{g/mol} \]

### Step 2: Calculate the Number of Moles of Na₂C₂O₄
To find the number of moles, we use the formula:
\[ \text{Number of moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} \]
\[ \text{Number of moles} = \frac{3.2 \, \text{g}}{134.00 \, \text{g/mol}} \approx 0.02388 \, \text{mol} \]

### Step 3: Determine the Equivalent Weight of Na₂C₂O₄
The equivalent weight is calculated using the formula:
\[ \text{Equivalent weight} = \frac{\text{Molar mass}}{n} \]
where \( n \) is the number of electrons transferred in the reaction. For oxalate ion (C₂O₄²⁻), it can lose 2 electrons during oxidation:
\[ n = 2 \]
Thus,
\[ \text{Equivalent weight} = \frac{134.00 \, \text{g/mol}}{2} = 67.00 \, \text{g/equiv} \]

### Step 4: Calculate the Number of Gram Equivalents
Using the formula:
\[ \text{Number of gram equivalents} = \frac{\text{mass (g)}}{\text{equivalent weight (g/equiv)}} \]
\[ \text{Number of gram equivalents} = \frac{3.2 \, \text{g}}{67.00 \, \text{g/equiv}} \approx 0.04776 \, \text{equiv} \]

### Step 5: Calculate the Normality of the Solution
Normality (N) is defined as the number of equivalents per liter of solution:
\[ \text{Normality} = \frac{\text{Number of equivalents}}{\text{Volume (L)}} \]
Since the volume is given in mL, we convert it to liters:
\[ 500 \, \text{mL} = 0.500 \, \text{L} \]
Now, substituting the values:
\[ \text{Normality} = \frac{0.04776 \, \text{equiv}}{0.500 \, \text{L}} \approx 0.09552 \, \text{N} \]

### Final Answer
The normality of the oxalate solution is approximately **0.096 N**.

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