22.7 mL of (N/10) `Na_(2)CO_(3)` solution neutralises 10.2 mL of a dilute `H_(2)SO_(4)` solution. The volume of water that must be added to 400 mL of this `H_(2)SO_(4)` solution in order to make it exactly N/10.

Options

A. 245 ml
B. 484.6 ml
C. 480 ml
D. 490.2 ml

Explanation

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

### Step 1: Determine the Normality of H₂SO₄
We know that the normality of Na₂CO₃ solution is N/10, and it neutralizes H₂SO₄. The balanced equation for the reaction is:

\[ \text{Na}_2\text{CO}_3 + \text{H}_2\text{SO}_4 \rightarrow \text{Na}_2\text{SO}_4 + \text{H}_2\text{O} + \text{CO}_2 \]

From the equation, we see that 1 mole of Na₂CO₃ reacts with 1 mole of H₂SO₄. Therefore, the equivalence can be set up as:

\[ N_{\text{Na}_2\text{CO}_3} \times V_{\text{Na}_2\text{CO}_3} = N_{\text{H}_2\text{SO}_4} \times V_{\text{H}_2\text{SO}_4} \]

Given:
- \( N_{\text{Na}_2\text{CO}_3} = \frac{1}{10} \, \text{N} \)
- \( V_{\text{Na}_2\text{CO}_3} = 22.7 \, \text{mL} \)
- \( V_{\text{H}_2\text{SO}_4} = 10.2 \, \text{mL} \)

Substituting the values:

\[ \frac{1}{10} \times 22.7 = N_{\text{H}_2\text{SO}_4} \times 10.2 \]

### Step 2: Calculate Normality of H₂SO₄
Rearranging the equation to find \( N_{\text{H}_2\text{SO}_4} \):

\[ N_{\text{H}_2\text{SO}_4} = \frac{\frac{1}{10} \times 22.7}{10.2} \]

Calculating:

\[ N_{\text{H}_2\text{SO}_4} = \frac{2.27}{10.2} \approx 0.222 \, \text{N} \]

### Step 3: Determine the Final Volume Required for N/10 H₂SO₄
We want to dilute this H₂SO₄ solution to N/10. Using the dilution formula:

\[ N_{\text{initial}} \times V_{\text{initial}} = N_{\text{final}} \times V_{\text{final}} \]

Where:
- \( N_{\text{initial}} = 0.222 \, \text{N} \)
- \( V_{\text{initial}} = 400 \, \text{mL} \)
- \( N_{\text{final}} = \frac{1}{10} \, \text{N} \)

Substituting the values:

\[ 0.222 \times 400 = \frac{1}{10} \times V_{\text{final}} \]

### Step 4: Solve for Final Volume
Rearranging gives:

\[ V_{\text{final}} = \frac{0.222 \times 400 \times 10}{1} \]

Calculating:

\[ V_{\text{final}} = 888 \, \text{mL} \]

### Step 5: Calculate the Volume of Water to be Added
The volume of water to be added is:

\[ V_{\text{water}} = V_{\text{final}} - V_{\text{initial}} = 888 \, \text{mL} - 400 \, \text{mL} = 488 \, \text{mL} \]

### Final Answer
The volume of water that must be added to 400 mL of the H₂SO₄ solution to make it exactly N/10 is approximately **488 mL**.

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