`KMnO_4` is prepared from the mineral pyrolusite, `MnO_2` (deep purple colour). It acts as an oxidising agent in the neutral, alkaline as well as acidic medium in acidic medium it is used in volumetric analysis for estimation of `Fe^(2+)`, `Cr_2O_4^(2-)` salts etc. The titrations are carried out in presence of `H_2SO_4`. However, before using it as a titrant, it is first standardised with standard oxalic acid solution or Mohr's salt solution . In one of the experiments on titration 26.8g of dry pure sodium oxalate `(Mw=123gmol^-1)` was dissolved in 1L of distilled water and then 100 mL of `2MH_2SO_4` were added. The solution was cooled. Now to this solution `0.1MKMnO_4` solution was added till a very faint pink colour persisted.
Q. If `0.1MK_2Cr_2O_7` solution were used in place of 0.1 `MKMnO_4` solution used would be

Options

A. 200 mL
B. 400 mL
C. 333.3 mL
D. 666.6 mL

Explanation

To solve the problem, we need to determine the volume of 0.1 M K2Cr2O7 solution required to react with the same amount of sodium oxalate as used with KMnO4. Here’s the step-by-step solution:

### Step 1: Calculate the moles of sodium oxalate (C2O4^2-)
Given:
- Mass of sodium oxalate = 26.8 g
- Molar mass of sodium oxalate (C2O4^2-) = 123 g/mol

\[ \text{Moles of sodium oxalate} = \frac{\text{mass}}{\text{molar mass}} = \frac{26.8 \, \text{g}}{123 \, \text{g/mol}} = 0.2187 \, \text{mol} \]

### Step 2: Determine the stoichiometry of the reaction
The balanced reaction between sodium oxalate and K2Cr2O7 in acidic medium is:

\[ \text{Cr}_2\text{O}_7^{2-} + 3 \text{C}_2\text{O}_4^{2-} \rightarrow 2 \text{Cr}^{3+} + 6 \text{CO}_2 + 7 \text{H}_2\text{O} \]

From the reaction, we see that 1 mole of K2Cr2O7 reacts with 3 moles of sodium oxalate.

### Step 3: Calculate the moles of K2Cr2O7 required
Using the stoichiometry from the balanced equation:

\[ \text{Moles of K2Cr2O7} = \frac{1}{3} \times \text{Moles of sodium oxalate} = \frac{1}{3} \times 0.2187 \, \text{mol} = 0.0729 \, \text{mol} \]

### Step 4: Calculate the volume of K2Cr2O7 solution needed
Given that the concentration of K2Cr2O7 is 0.1 M, we can use the formula:

\[ \text{Volume (L)} = \frac{\text{Moles}}{\text{Concentration (M)}} \]

Substituting the values:

\[ \text{Volume (L)} = \frac{0.0729 \, \text{mol}}{0.1 \, \text{mol/L}} = 0.729 \, \text{L} \]

Converting to mL:

\[ \text{Volume (mL)} = 0.729 \, \text{L} \times 1000 \, \text{mL/L} = 729 \, \text{mL} \]

### Final Answer
If 0.1 M K2Cr2O7 solution were used instead of 0.1 M KMnO4 solution, the volume required would be **729 mL**.

---