Match the following`{:(,"Column -I",,"Column -II"),((A),H_(3)PO_(3),(p),m=2.79" mol/kg"),((B),20 " vol . of "H_(2)O_(2),(q),M=2.31" mol"L^(-1)),((C),"3 M NaCl solution with "d=1.25 g cm^(-3),(r),1.7 " mol "L^(-1)),((D),"Mole fraction of ethanol in" H_(2)O = 0.04 ,(s),"Dibasic "),(,,(t),3.4 N):}`

Answer

To solve the problem of matching the items in Column I with those in Column II, we will analyze each item in Column I and determine the corresponding value from Column II step by step.

### Step 1: Analyze H₃PO₃ (Phosphorous Acid)
- **Basicity**: H₃PO₃ has two acidic hydrogens, making it dibasic.
- **Match**: Therefore, H₃PO₃ corresponds to option (s) "Dibasic".

### Step 2: Analyze 20 Vol. of H₂O₂ (Hydrogen Peroxide)
- **Volume Strength**: 20 volume means that 1 liter of H₂O₂ can release 20 liters of O₂.
- **Molarity Calculation**:
- Molarity (M) = Volume Strength / 11.2 = 20 / 11.2 = 1.79 mol/L (approximately).
- **Normality Calculation**:
- Normality (N) = Volume Strength / 5.6 = 20 / 5.6 = 3.57 N (approximately).
- **Match**: Therefore, 20 vol. of H₂O₂ corresponds to (r) "1.7 mol L⁻¹" and (t) "3.4 N".

### Step 3: Analyze 3 M NaCl Solution with d = 1.25 g/cm³
- **Density**: Given density is 1.25 g/cm³, which is equivalent to 1.25 g/mL.
- **Weight of Solution**: For 1 L of solution, the weight = 1 L × 1.25 g/mL = 1250 g.
- **Weight of Solute**: Molarity (3 M) means there are 3 moles of NaCl.
- Weight of NaCl = 3 moles × 58.5 g/mol = 175.5 g.
- **Weight of Solvent**: Weight of solvent = Weight of solution - Weight of solute = 1250 g - 175.5 g = 1074.5 g = 1.0745 kg.
- **Molality Calculation**:
- Molality (m) = moles of solute / kg of solvent = 3 moles / 1.0745 kg ≈ 2.79 mol/kg.
- **Match**: Therefore, this corresponds to (p) "m = 2.79 mol/kg".

### Step 4: Analyze Mole Fraction of Ethanol in H₂O = 0.04
- **Mole Fraction**: Given mole fraction of ethanol is 0.04, thus moles of water = 1 - 0.04 = 0.96.
- **Weight of Water**:
- Weight of water = 0.96 moles × 18 g/mol = 17.28 g.
- **Volume of Water**: Volume of water = 17.28 g / 1 g/mL = 17.28 mL = 0.01728 L.
- **Molarity Calculation**:
- Molarity = moles of ethanol / volume of solution in L = 0.04 / 0.01728 ≈ 2.31 mol/L.
- **Match**: Therefore, this corresponds to (q) "M = 2.31 mol L⁻¹".

### Final Matching:
- A → s (H₃PO₃ is dibasic)
- B → r (20 vol. of H₂O₂ is 1.7 mol L⁻¹)
- C → p (3 M NaCl solution has m = 2.79 mol/kg)
- D → q (Mole fraction of ethanol gives M = 2.31 mol L⁻¹)

### Summary of Matches:
- A - s
- B - r
- C - p
- D - q