`4.0 g` of a mixture of `Nacl` and `Na_(2) CO_(3)` was dissolved in water and volume made up to `250 mL`. `25 mL` of this solution required `50 mL` of `N//10 HCl` for complete neutralisation. Calculate the percentage composition of the original mixture.

Explanation

To solve the problem, we need to determine the percentage composition of a mixture of NaCl and Na2CO3 based on the given data. Here’s a step-by-step breakdown of the solution:

### Step 1: Define Variables
Let:
- \( A \) = mass of NaCl (in grams)
- \( B \) = mass of Na2CO3 (in grams)

From the problem, we know:
\[ A + B = 4.0 \, \text{g} \] (Equation 1)

### Step 2: Determine the Reaction and Normality
NaCl does not react with HCl as it is a neutral salt, while Na2CO3 will react with HCl. The reaction can be represented as:
\[ \text{Na}_2\text{CO}_3 + 2 \text{HCl} \rightarrow 2 \text{NaCl} + \text{H}_2\text{O} + \text{CO}_2 \]

### Step 3: Calculate Milliequivalents of HCl Used
The problem states that 25 mL of the solution required 50 mL of N/10 HCl for complete neutralization.

First, calculate the milliequivalents of HCl:
- Normality of HCl = \( \frac{1}{10} \, \text{N} \)
- Volume of HCl used = \( 50 \, \text{mL} \)

Using the formula for milliequivalents:
\[ \text{Milliequivalents of HCl} = \text{Normality} \times \text{Volume (mL)} = \frac{1}{10} \times 50 = 5 \, \text{milliequivalents} \]

Since this is for 25 mL of the solution, for 250 mL, the total milliequivalents required will be:
\[ \text{Total milliequivalents} = 5 \times \frac{250}{25} = 50 \, \text{milliequivalents} \]

### Step 4: Relate Milliequivalents to Na2CO3
The milliequivalents of Na2CO3 can be calculated using the formula:
\[ \text{Milliequivalents} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} \times \text{n factor} \]
For Na2CO3:
- Molar mass = \( 23 \times 2 + 12 + 16 \times 3 = 106 \, \text{g/mol} \)
- n factor = 2 (because 1 mole of Na2CO3 gives 2 equivalents of Na+)

Setting up the equation:
\[ 50 = \frac{B}{106} \times 2 \]
Solving for \( B \):
\[ B = \frac{50 \times 106}{2} = 2650 \, \text{mg} = 2.65 \, \text{g} \]

### Step 5: Calculate Mass of NaCl
Using Equation 1:
\[ A + B = 4.0 \, \text{g} \]
Substituting \( B \):
\[ A + 2.65 = 4.0 \]
\[ A = 4.0 - 2.65 = 1.35 \, \text{g} \]

### Step 6: Calculate Percentage Composition
Now, we can calculate the percentage composition of each component in the mixture:
- Percentage of Na2CO3:
\[ \text{Percentage of Na2CO3} = \left( \frac{B}{4.0} \right) \times 100 = \left( \frac{2.65}{4.0} \right) \times 100 = 66.25\% \]

- Percentage of NaCl:
\[ \text{Percentage of NaCl} = \left( \frac{A}{4.0} \right) \times 100 = \left( \frac{1.35}{4.0} \right) \times 100 = 33.75\% \]

### Final Answer
- Percentage of NaCl = 33.75%
- Percentage of Na2CO3 = 66.25%