The density of a `"3 M Na_(2)S_(2)O_(3)` (sodium thiosulphate) solution is `"1.25 g cm"^(-3)`. Calculate (iii) the molality of `Na^(+)` and `S_(2)O_(3)^(2-)` ions.

Explanation

To calculate the molality of \( \text{Na}^+ \) and \( \text{S}_2\text{O}_3^{2-} \) ions in a 3 M sodium thiosulfate solution with a density of 1.25 g/cm³, follow these steps:

### Step 1: Calculate the mass of sodium thiosulfate in the solution
A 3 M solution means there are 3 moles of sodium thiosulfate (\( \text{Na}_2\text{S}_2\text{O}_3 \)) in 1 liter of solution.

**Molecular weight of \( \text{Na}_2\text{S}_2\text{O}_3 \)**:
- Sodium (Na) = 23 g/mol
- Sulfur (S) = 32 g/mol
- Oxygen (O) = 16 g/mol

\[ \text{Molecular weight} = 2(23) + 2(32) + 3(16) = 46 + 64 + 48 = 158 \text{ g/mol} \]

**Mass of solute**:
\[ \text{Mass of solute} = \text{Number of moles} \times \text{Molecular weight} = 3 \text{ moles} \times 158 \text{ g/mol} = 474 \text{ g} \]

### Step 2: Calculate the total mass of the solution
Using the density of the solution, we can find the total mass of the solution.

**Density** = 1.25 g/cm³, and since 1 liter = 1000 cm³:
\[ \text{Mass of solution} = \text{Density} \times \text{Volume} = 1.25 \text{ g/cm}^3 \times 1000 \text{ cm}^3 = 1250 \text{ g} \]

### Step 3: Calculate the mass of the solvent
The mass of the solvent (water) can be calculated by subtracting the mass of the solute from the total mass of the solution.

\[ \text{Mass of solvent} = \text{Mass of solution} - \text{Mass of solute} = 1250 \text{ g} - 474 \text{ g} = 776 \text{ g} \]

Convert this mass into kilograms for the molality calculation:
\[ \text{Mass of solvent in kg} = \frac{776 \text{ g}}{1000} = 0.776 \text{ kg} \]

### Step 4: Calculate the molality of the solution
The formula for molality (\( m \)) is given by:
\[ m = \frac{\text{Number of moles of solute}}{\text{Mass of solvent in kg}} \]

Substituting the values:
\[ m = \frac{3 \text{ moles}}{0.776 \text{ kg}} \approx 3.86 \text{ mol/kg} \]

### Step 5: Determine the molality of ions
The dissociation of sodium thiosulfate in solution is:
\[ \text{Na}_2\text{S}_2\text{O}_3 \rightarrow 2 \text{Na}^+ + \text{S}_2\text{O}_3^{2-} \]

From the dissociation:
- 1 mole of \( \text{Na}_2\text{S}_2\text{O}_3 \) produces 2 moles of \( \text{Na}^+ \) ions.
- 1 mole of \( \text{Na}_2\text{S}_2\text{O}_3 \) produces 1 mole of \( \text{S}_2\text{O}_3^{2-} \) ions.

Thus:
- The molality of \( \text{Na}^+ \) ions = \( 2 \times 3.86 \approx 7.72 \text{ mol/kg} \)
- The molality of \( \text{S}_2\text{O}_3^{2-} \) ions = \( 3.86 \text{ mol/kg} \)

### Final Results
- Molality of \( \text{Na}^+ \) ions = 7.72 mol/kg
- Molality of \( \text{S}_2\text{O}_3^{2-} \) ions = 3.86 mol/kg