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 dry anhydrous sodium oxalate (Na₂C₂O₄) in distilled water to make a 500 mL solution, we will follow these steps:

### Step 1: Calculate the number of moles of Na₂C₂O₄

First, we need to find the molar mass of sodium oxalate (Na₂C₂O₄). The molar mass can be calculated as follows:

- Sodium (Na): 23 g/mol × 2 = 46 g/mol
- Carbon (C): 12 g/mol × 2 = 24 g/mol
- Oxygen (O): 16 g/mol × 4 = 64 g/mol

Adding these together gives:

\[ \text{Molar mass of Na₂C₂O₄} = 46 + 24 + 64 = 134 \text{ g/mol} \]

Now, we can calculate the number of moles of Na₂C₂O₄ in 3.2 g:

\[ \text{Number of moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} = \frac{3.2 \text{ g}}{134 \text{ g/mol}} \approx 0.02388 \text{ moles} \]

### Step 2: Calculate the molarity of the solution

Molarity (M) is defined as the number of moles of solute per liter of solution. Since the solution is made up to 500 mL, we need to convert this volume to liters:

\[ 500 \text{ mL} = 0.500 \text{ L} \]

Now, we can calculate the molarity:

\[ \text{Molarity (M)} = \frac{\text{number of moles}}{\text{volume (L)}} = \frac{0.02388 \text{ moles}}{0.500 \text{ L}} \approx 0.04776 \text{ M} \]

### Step 3: Determine the n-factor for oxalate ion

The n-factor for oxalate (C₂O₄²⁻) can be determined by looking at the change in oxidation state. In the reaction, oxalate is oxidized to carbon dioxide (CO₂).

- In C₂O₄²⁻, the oxidation state of carbon is +3.
- In CO₂, the oxidation state of carbon is +4.

The change in oxidation state for one carbon atom is:

\[ \text{Change} = +4 - (+3) = +1 \]

Since there are 2 carbon atoms in oxalate, the total change in oxidation state is:

\[ \text{Total change} = 2 \times 1 = 2 \]

Thus, the n-factor for oxalate is 2.

### Step 4: Calculate the normality of the solution

Normality (N) is related to molarity (M) by the equation:

\[ \text{Normality (N)} = \text{Molarity (M)} \times \text{n-factor} \]

Substituting the values we have:

\[ \text{Normality} = 0.04776 \text{ M} \times 2 \approx 0.09552 \text{ N} \]

Rounding this to three significant figures gives us:

\[ \text{Normality} \approx 0.096 \text{ N} \]

### Final Answer:

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

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