A sample of sodium carbonate contains impurity of sodium sulphate. 1.25 g of this sample are dissolved in water and volume made up to 250 mL. 25 mL of this solution neutralise 20 mL of `N/10` sulphuric acid.
Calculate the percentage of sodium carbonate in the sample.

Answer

To solve the problem step by step, we will follow these calculations:

### Step 1: Determine the equivalent weight of sodium carbonate (Na2CO3)
The formula for sodium carbonate is Na2CO3. The molar mass of Na2CO3 can be calculated as follows:
- Sodium (Na) = 23 g/mol, and there are 2 sodium atoms: 2 × 23 = 46 g/mol
- Carbon (C) = 12 g/mol
- Oxygen (O) = 16 g/mol, and there are 3 oxygen atoms: 3 × 16 = 48 g/mol

Total molar mass of Na2CO3 = 46 + 12 + 48 = 106 g/mol

The equivalent weight of Na2CO3 is given by:
\[ \text{Equivalent weight} = \frac{\text{Molar mass}}{\text{n (valency)}} = \frac{106}{2} = 53 \text{ g/equiv} \]

### Step 2: Calculate the amount of Na2CO3 that reacts with the sulfuric acid
We know that 25 mL of the sample solution neutralizes 20 mL of N/10 sulfuric acid. First, we need to find out how much Na2CO3 is in 250 mL of the sample solution.

Since 25 mL of the sample solution neutralizes 20 mL of N/10 sulfuric acid, we can scale this up to find out how much 250 mL of the sample solution would neutralize:
\[ \text{Volume of sample solution} = 250 \text{ mL} \]
\[ \text{Volume of sulfuric acid} = 20 \text{ mL} \times \frac{250}{25} = 200 \text{ mL} \]

### Step 3: Calculate the amount of Na2CO3 using the normality formula
Using the formula:
\[ \text{Amount of Na2CO3} = N \times E \times V / 1000 \]
Where:
- \(N\) = Normality of sulfuric acid = 0.1 N (N/10)
- \(E\) = Equivalent weight of Na2CO3 = 53 g/equiv
- \(V\) = Volume of sulfuric acid = 200 mL

Substituting the values:
\[ \text{Amount of Na2CO3} = 0.1 \times 53 \times 200 / 1000 \]
\[ = 1.06 \text{ g} \]

### Step 4: Calculate the percentage of sodium carbonate in the sample
The total weight of the sample is 1.25 g. Now we can find the percentage of Na2CO3 in the sample:
\[ \text{Percentage of Na2CO3} = \left(\frac{\text{Amount of Na2CO3}}{\text{Total weight of sample}}\right) \times 100 \]
\[ = \left(\frac{1.06}{1.25}\right) \times 100 \]
\[ = 84.8\% \]

### Final Answer
The percentage of sodium carbonate in the sample is **84.8%**.

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