If 1.34g `Na_(2)C_(2)O_(4)`is dissolved in 500 mL of water and this solution is titrated with `M/10 KMnO_(4)` solution in acidic medium, the volume of`KMnO_(4)` used is :

Options

A. 200mL
B. `200/2mL`
C. 40mL
D. none of these

Explanation

To solve the problem, we need to follow these steps:

### Step 1: Calculate the Molar Mass of Sodium Oxalate (Na₂C₂O₄)

The molar mass of Na₂C₂O₄ 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:
\[ \text{Molar mass of Na}_2\text{C}_2\text{O}_4 = 46 + 24 + 64 = 134 \text{ g/mol} \]

### Step 2: Calculate the Number of Moles of Na₂C₂O₄

Using the mass of Na₂C₂O₄ given (1.34 g), we can find the number of moles:
\[ \text{Number of moles} = \frac{\text{mass}}{\text{molar mass}} = \frac{1.34 \text{ g}}{134 \text{ g/mol}} = 0.01 \text{ mol} \]

### Step 3: Determine the Valency Factor (n)

In the reaction, oxalate ion (C₂O₄²⁻) is oxidized to carbon dioxide (CO₂), and permanganate ion (MnO₄⁻) is reduced to manganese ion (Mn²⁺). The change in oxidation states is:
- Carbon in C₂O₄²⁻ goes from +3 to +4 (total change of +2 for 2 carbon atoms).
- Manganese in MnO₄⁻ goes from +7 to +2 (total change of -5).

Thus, the valency factor (n) for the reaction is 5.

### Step 4: Calculate the Equivalent of Na₂C₂O₄

The equivalent of Na₂C₂O₄ can be calculated as:
\[ \text{Equivalent} = \text{Number of moles} \times n = 0.01 \text{ mol} \times 5 = 0.05 \text{ equivalents} \]

### Step 5: Use the Normality of KMnO₄ to Find the Volume Used

The normality of KMnO₄ solution is given as M/10, which is equivalent to 0.1 N. We can use the formula:
\[ \text{Equivalent} = \text{Normality} \times \text{Volume (L)} \]
Rearranging gives:
\[ \text{Volume (L)} = \frac{\text{Equivalent}}{\text{Normality}} = \frac{0.05 \text{ equivalents}}{0.1 \text{ N}} = 0.5 \text{ L} \]

### Step 6: Convert Volume to mL

Since the volume is required in mL:
\[ \text{Volume (mL)} = 0.5 \text{ L} \times 1000 = 500 \text{ mL} \]

### Final Answer

The volume of KMnO₄ used is **500 mL**.

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