The density of 3 M aqueous solution of sodium thiosulphate is 1.25 g/mL. Calculate (i) mole fraction of sodium thiosulphate (ii) molalities of `Na^+` and `S_(2)O_(3)^(2-)` ions.

Answer

To solve the problem, we will follow these steps:

### Given Data:
- Molarity of sodium thiosulfate (Na2S2O3) = 3 M
- Density of the solution = 1.25 g/mL

### Step 1: Calculate the mass of sodium thiosulfate in 1 liter of solution.
- Molar mass of Na2S2O3 = (2 × 23) + (2 × 32) + (3 × 16) = 46 + 64 + 48 = 158 g/mol
- Moles of Na2S2O3 in 1 L = 3 moles
- Mass of Na2S2O3 = moles × molar mass = 3 moles × 158 g/mol = 474 g

### Step 2: Calculate the mass of the solution.
- Volume of the solution = 1 L = 1000 mL
- Mass of the solution = volume × density = 1000 mL × 1.25 g/mL = 1250 g

### Step 3: Calculate the mass of the solvent (water).
- Mass of solvent (water) = mass of solution - mass of solute
- Mass of solvent = 1250 g - 474 g = 776 g

### Step 4: Calculate the mole fraction of sodium thiosulfate.
- Moles of water = mass of water / molar mass of water = 776 g / 18 g/mol = 43.11 moles
- Mole fraction of Na2S2O3 = moles of Na2S2O3 / (moles of Na2S2O3 + moles of water)
- Mole fraction of Na2S2O3 = 3 / (3 + 43.11) = 3 / 46.11 ≈ 0.065

### Step 5: Calculate the molalities of Na+ and S2O3^2- ions.
- Sodium thiosulfate dissociates as: Na2S2O3 → 2 Na+ + S2O3^2-
- Moles of Na+ = 2 × moles of Na2S2O3 = 2 × 3 = 6 moles
- Moles of S2O3^2- = moles of Na2S2O3 = 3 moles

#### For Na+:
- Molality of Na+ = moles of Na+ / (mass of solvent in kg)
- Mass of solvent in kg = 776 g / 1000 = 0.776 kg
- Molality of Na+ = 6 moles / 0.776 kg ≈ 7.732 mol/kg

#### For S2O3^2-:
- Molality of S2O3^2- = moles of S2O3^2- / (mass of solvent in kg)
- Molality of S2O3^2- = 3 moles / 0.776 kg ≈ 3.866 mol/kg

### Final Answers:
1. Mole fraction of sodium thiosulfate (Na2S2O3) ≈ 0.065
2. Molality of Na+ ≈ 7.732 mol/kg
3. Molality of S2O3^2- ≈ 3.866 mol/kg