The density of a `"3 M Na_(2)S_(2)O_(3)` (sodium thiosulphate) solution is `"1.25 g cm"^(-3)`. Calculate (ii) the mole fraction of sodium thiosulphate

Explanation

To calculate the mole fraction of sodium thiosulfate (Na₂S₂O₃) in a 3 M solution with a density of 1.25 g/cm³, we will follow these steps:

### Step 1: Calculate the mass of sodium thiosulfate in the solution
Given that the solution is 3 M, this means there are 3 moles of sodium thiosulfate in 1 liter of solution.

- **Molecular weight of Na₂S₂O₃**:
- Na: 23 g/mol (2 Na = 46 g/mol)
- S: 32 g/mol (2 S = 64 g/mol)
- O: 16 g/mol (3 O = 48 g/mol)

Total = 46 + 64 + 48 = **158 g/mol**

- **Mass of sodium thiosulfate**:
\[ \text{Mass} = \text{Number of moles} \times \text{Molecular weight} = 3 \, \text{moles} \times 158 \, \text{g/mol} = 474 \, \text{grams} \]

### Step 2: Calculate the mass of the solution
Using the density of the solution to find the mass:
- **Density** = 1.25 g/cm³
- **Volume** = 1 L = 1000 cm³

\[ \text{Mass of solution} = \text{Density} \times \text{Volume} = 1.25 \, \text{g/cm}^3 \times 1000 \, \text{cm}^3 = 1250 \, \text{grams} \]

### Step 3: Calculate the mass of the solvent (water)
\[ \text{Mass of solvent} = \text{Mass of solution} - \text{Mass of solute} = 1250 \, \text{grams} - 474 \, \text{grams} = 776 \, \text{grams} \]

### Step 4: Calculate the number of moles of the solvent (water)
- **Molecular weight of water (H₂O)**: 18 g/mol

\[ \text{Number of moles of solvent} = \frac{\text{Mass of solvent}}{\text{Molecular weight of water}} = \frac{776 \, \text{grams}}{18 \, \text{g/mol}} \approx 43.11 \, \text{moles} \]

### Step 5: Calculate the mole fraction of sodium thiosulfate
The mole fraction (X) is given by the formula:
\[ X_{\text{Na}_2\text{S}_2\text{O}_3} = \frac{\text{Moles of Na}_2\text{S}_2\text{O}_3}{\text{Moles of Na}_2\text{S}_2\text{O}_3 + \text{Moles of solvent}} \]

Substituting the values:
\[ X_{\text{Na}_2\text{S}_2\text{O}_3} = \frac{3}{3 + 43.11} = \frac{3}{46.11} \approx 0.065 \]

### Final Answer
The mole fraction of sodium thiosulfate in the solution is approximately **0.065**.

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