STATEMENT-1: N/10, 100 ml `KMnO_(4)` solution is sufficient to oxidise M/10, 50 ml `FeC_(2)O_(4)` solution in acidic medium.
STATEMENT-2: The left solution of statement -1 is sufficient to react with 8.33 ml of M/10 `K_(2)Cr_(2)O_(7)` solution in acidic medium.
STATEMENT-3: 1.06 g `Na_(2)CO_(3)` will require 100 ml of 0.1 M HCl solution with phenolphthalein.

Options

A. F T T
B. F T F
C. T F T
D. T T F

Answer

To solve the question regarding the statements about redox reactions, we will analyze each statement step by step.

### Step 1: Analyze Statement 1
**Statement 1:** N/10, 100 ml `KMnO4` solution is sufficient to oxidize M/10, 50 ml `FeC2O4` solution in acidic medium.

1. **Calculate the equivalents of `KMnO4`:**
- Normality of `KMnO4` = N/10 = 0.1 N
- Volume of `KMnO4` = 100 ml = 0.1 L
- Equivalents of `KMnO4` = Normality × Volume (in L) = 0.1 × 0.1 = 0.01 equivalents.

2. **Calculate the equivalents of `FeC2O4`:**
- Molarity of `FeC2O4` = M/10 = 0.1 M
- Volume of `FeC2O4` = 50 ml = 0.05 L
- Valency factor of `FeC2O4`:
- Iron changes from +2 to +3 (change of 1).
- Each carbon in `C2O4` changes from +3 to +4 (change of 1 per carbon, total change of 2 for 2 carbons).
- Total change = 1 (Fe) + 2 (C) = 3.
- Equivalents of `FeC2O4` = Molarity × Volume × Valency Factor = 0.1 × 0.05 × 3 = 0.015 equivalents.

3. **Comparison:**
- `KMnO4` has 0.01 equivalents and `FeC2O4` has 0.015 equivalents. Since `KMnO4` has fewer equivalents, it is not sufficient to oxidize `FeC2O4`.

**Conclusion for Statement 1:** False.

### Step 2: Analyze Statement 2
**Statement 2:** The left solution of statement 1 is sufficient to react with 8.33 ml of M/10 `K2Cr2O7` solution in acidic medium.

1. **Calculate the remaining equivalents of `FeC2O4`:**
- From Statement 1, `FeC2O4` had 0.015 equivalents, and `KMnO4` provided 0.01 equivalents.
- Remaining equivalents of `FeC2O4` = 0.015 - 0.01 = 0.005 equivalents.

2. **Calculate the equivalents of `K2Cr2O7`:**
- Molarity of `K2Cr2O7` = M/10 = 0.1 M
- Volume of `K2Cr2O7` = 8.33 ml = 0.00833 L
- Valency factor of `K2Cr2O7`:
- Chromium changes from +6 to +3 (change of 3).
- Total change = 6 (for 1 Cr) × 1 = 6.
- Equivalents of `K2Cr2O7` = Molarity × Volume × Valency Factor = 0.1 × 0.00833 × 6 = 0.005 equivalents.

3. **Comparison:**
- The remaining equivalents of `FeC2O4` (0.005) matches the equivalents of `K2Cr2O7` (0.005), indicating they can react completely.

**Conclusion for Statement 2:** True.

### Step 3: Analyze Statement 3
**Statement 3:** 1.06 g `Na2CO3` will require 100 ml of 0.1 M HCl solution with phenolphthalein.

1. **Calculate the equivalents of `Na2CO3`:**
- Molar mass of `Na2CO3` = 106 g/mol.
- Moles of `Na2CO3` = Mass / Molar Mass = 1.06 g / 106 g/mol = 0.01 moles.
- Valency factor of `Na2CO3` = 2 (as it can donate 2 protons).
- Equivalents of `Na2CO3` = Moles × Valency Factor = 0.01 × 2 = 0.02 equivalents.

2. **Calculate the equivalents of `HCl`:**
- Molarity of `HCl` = 0.1 M
- Volume of `HCl` = 100 ml = 0.1 L
- Valency factor of `HCl` = 1.
- Equivalents of `HCl` = Molarity × Volume × Valency Factor = 0.1 × 0.1 × 1 = 0.01 equivalents.

3. **Comparison:**
- `Na2CO3` requires 0.02 equivalents, but `HCl` provides only 0.01 equivalents, which is insufficient.

**Conclusion for Statement 3:** False.

### Final Conclusions:
- Statement 1: False
- Statement 2: True
- Statement 3: False

### Summary of Results:
- Statement 1: F
- Statement 2: T
- Statement 3: F