Calculate the amount of `95%` pure `Na_(2)CO_(3)` required to prepare `5` litre of `0.5 M` solution.

Explanation

To calculate the amount of 95% pure Na₂CO₃ required to prepare a 5-liter solution of 0.5 M concentration, follow these steps:

### Step 1: Calculate the number of moles of Na₂CO₃ needed
The formula to calculate the number of moles (n) is:
\[ n = M \times V \]
where:
- \( M \) = molarity of the solution (in moles per liter)
- \( V \) = volume of the solution (in liters)

Given:
- \( M = 0.5 \, \text{mol/L} \)
- \( V = 5 \, \text{L} \)

Substituting the values:
\[ n = 0.5 \, \text{mol/L} \times 5 \, \text{L} = 2.5 \, \text{moles} \]

### Step 2: Calculate the mass of Na₂CO₃ required for 100% purity
To find the mass (m) of Na₂CO₃, use the formula:
\[ m = n \times \text{molar mass} \]
The molar mass of Na₂CO₃ (sodium carbonate) is approximately 106 g/mol.

Substituting the values:
\[ m = 2.5 \, \text{moles} \times 106 \, \text{g/mol} = 265 \, \text{grams} \]

### Step 3: Adjust for the purity of Na₂CO₃
Since we have 95% pure Na₂CO₃, we need to find the mass of the pure substance that corresponds to the 265 grams calculated for 100% purity.

To find the mass of the 95% pure Na₂CO₃ required, use the formula:
\[ \text{mass of 95% Na₂CO₃} = \frac{\text{mass of pure Na₂CO₃}}{\text{purity fraction}} \]
The purity fraction for 95% is 0.95.

Substituting the values:
\[ \text{mass of 95% Na₂CO₃} = \frac{265 \, \text{grams}}{0.95} \approx 278.95 \, \text{grams} \]

### Step 4: Round off the answer
Rounding off to a reasonable number of significant figures, we find:
\[ \text{mass of 95% Na₂CO₃} \approx 279 \, \text{grams} \]

### Final Answer
To prepare 5 liters of a 0.5 M Na₂CO₃ solution, you will need approximately **279 grams** of 95% pure Na₂CO₃.

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