Calculate the formality of sodium thiosulphate `(Na_(2)S_(2)O_(3).5H_(2)O)` solution , 1.24 g of which are dissolved in `100 cm^(3)` of the solution.

Explanation

To calculate the formality of sodium thiosulphate (Na₂S₂O₃·5H₂O) in a solution, we will follow these steps:

### Step 1: Understand the Concept of Formality
Formality is defined as the number of formula units of solute (in this case, Na₂S₂O₃·5H₂O) dissolved per liter of solution. It is similar to molarity but is used for ionic compounds in solution.

### Step 2: Identify the Given Data
- Mass of Na₂S₂O₃·5H₂O = 1.24 g
- Volume of solution = 100 cm³ = 100 mL = 0.1 L (since 1000 mL = 1 L)

### Step 3: Calculate the Molar Mass of Na₂S₂O₃·5H₂O
To calculate the molar mass (Mₘ) of sodium thiosulphate pentahydrate:
- Sodium (Na): 22.99 g/mol × 2 = 45.98 g/mol
- Sulfur (S): 32.07 g/mol × 2 = 64.14 g/mol
- Oxygen (O): 16.00 g/mol × 3 = 48.00 g/mol
- Water (H₂O): 18.02 g/mol × 5 = 90.10 g/mol

Now, add these values together:
\[ M_m = 45.98 + 64.14 + 48.00 + 90.10 = 248.22 \, \text{g/mol} \]

### Step 4: Calculate the Number of Moles of Na₂S₂O₃·5H₂O
Using the formula:
\[ \text{Number of moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} \]
Substituting the values:
\[ \text{Number of moles} = \frac{1.24 \, \text{g}}{248.22 \, \text{g/mol}} \approx 0.00499 \, \text{mol} \]

### Step 5: Calculate the Formality
Using the formula for formality:
\[ \text{Formality} = \frac{\text{Number of moles}}{\text{Volume of solution (L)}} \]
Substituting the values:
\[ \text{Formality} = \frac{0.00499 \, \text{mol}}{0.1 \, \text{L}} = 0.0499 \, \text{F} \approx 0.05 \, \text{F} \]

### Final Answer
The formality of the sodium thiosulphate solution is approximately **0.05 F**.

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