Three different solutions of oxidising agents `KMnO_(4),K_(2)Cr_(2)O_(7) " and "I_(2)` is titrated separately with 0.158 gm of `Na_(2)S_(2)O_(3)`. If molarity of each oxidising agent is 0.1 M and reactions are :
I.`MnO_(4)^(-)+S_(2)O_(3)^(2-) toMnO_(2)+SO_(4)^(2-)`
II `CrO_(7)^(2-)+S_(2)O_(3)^(2-) to Cr^(3+)+SO_(4)^(2-)`
III. `I_(2)+S_(2)O_(3)^(2-) to S_(4)O_(6)^(2-)+I^(-)`

Options

A. Volume of `KMnO_(4)` used is maximum
B. Volume of iodine used is minimum
C. Number of equivalents of `I_(2)` used in titration is maximum
D. Gram equivalent of `Na_(2)S_(2)O_(3)` are same in all the reactions.

Answer

To solve the problem, we need to determine the volumes of the three oxidizing agents (KMnO4, K2Cr2O7, and I2) that react with a given mass of sodium thiosulfate (Na2S2O3). We will also calculate the equivalents of each oxidizing agent based on the balanced reactions provided.

### Step 1: Calculate the number of moles of Na2S2O3

Given:
- Mass of Na2S2O3 = 0.158 g
- Molar mass of Na2S2O3 = 158 g/mol

Number of moles of Na2S2O3 = \(\frac{\text{mass}}{\text{molar mass}} = \frac{0.158 \, \text{g}}{158 \, \text{g/mol}} = 0.001 \, \text{mol}\)

**Hint:** To find the number of moles, divide the mass of the substance by its molar mass.

### Step 2: Determine the equivalents of Na2S2O3

The reaction of sodium thiosulfate (Na2S2O3) with oxidizing agents indicates that it can donate 2 electrons per molecule (since it changes from S2O3^2- to S4O6^2-). Therefore, the equivalent factor for Na2S2O3 is 2.

Equivalents of Na2S2O3 = Number of moles × Equivalent factor = \(0.001 \, \text{mol} \times 2 = 0.002 \, \text{equivalents}\)

**Hint:** The equivalent factor is the number of electrons transferred in the reaction. Multiply the number of moles by this factor to find equivalents.

### Step 3: Calculate the equivalents of each oxidizing agent

1. **For KMnO4:**
- The balanced reaction shows that 1 mole of KMnO4 reacts with 5 moles of S2O3^2-.
- Therefore, the equivalent factor for KMnO4 is 5.

Equivalents of KMnO4 = Molarity × Volume × Equivalent factor
Let the volume of KMnO4 be \(V_1\) L.
\[ 0.1 \, \text{mol/L} \times V_1 \, \text{L} \times 5 = 0.002 \, \text{equivalents} \]
\[ V_1 = \frac{0.002}{0.1 \times 5} = 0.004 \, \text{L} = 4 \, \text{mL} \]

2. **For K2Cr2O7:**
- The balanced reaction shows that 1 mole of K2Cr2O7 reacts with 6 moles of S2O3^2-.
- Therefore, the equivalent factor for K2Cr2O7 is 6.

Let the volume of K2Cr2O7 be \(V_2\) L.
\[ 0.1 \, \text{mol/L} \times V_2 \, \text{L} \times 6 = 0.002 \, \text{equivalents} \]
\[ V_2 = \frac{0.002}{0.1 \times 6} = 0.00333 \, \text{L} = 3.33 \, \text{mL} \]

3. **For I2:**
- The balanced reaction shows that 1 mole of I2 reacts with 2 moles of S2O3^2-.
- Therefore, the equivalent factor for I2 is 2.

Let the volume of I2 be \(V_3\) L.
\[ 0.1 \, \text{mol/L} \times V_3 \, \text{L} \times 2 = 0.002 \, \text{equivalents} \]
\[ V_3 = \frac{0.002}{0.1 \times 2} = 0.01 \, \text{L} = 10 \, \text{mL} \]

### Summary of Volumes
- Volume of KMnO4 = 4 mL
- Volume of K2Cr2O7 = 3.33 mL
- Volume of I2 = 10 mL

### Conclusion
- The maximum volume used is for I2 (10 mL), and the minimum is for K2Cr2O7 (3.33 mL).