Battery acid `(H_(2)SO_(4))` has density 1.285 g `cm^(-3)` , 10.0 `cm^(3)` of this acid is diluted to 1L . 25.0 `cm^(3)` of this diluted solution requires `25.0 cm^(3)` of 0.1 N sodium hydroxide solution for neutralization. The percentage of sulphuric acid by mass in the battery acid is :

Options

A. 98
B. 38
C. 19
D. 49

Answer

To solve the problem step by step, we need to calculate the percentage of sulfuric acid by mass in the battery acid. Here’s how to do it:

### Step 1: Write the Neutralization Reaction
The first step is to write the balanced chemical equation for the neutralization reaction between sulfuric acid (H₂SO₄) and sodium hydroxide (NaOH).

**Reaction:**
\[ \text{H}_2\text{SO}_4 + 2 \text{NaOH} \rightarrow \text{Na}_2\text{SO}_4 + 2 \text{H}_2\text{O} \]

### Step 2: Calculate Moles of NaOH
We know that 25.0 cm³ (or 0.025 L) of 0.1 N NaOH is used for neutralization.

**Moles of NaOH:**
\[ \text{Moles of NaOH} = \text{Normality} \times \text{Volume (L)} = 0.1 \, \text{N} \times 0.025 \, \text{L} = 0.0025 \, \text{moles} \]

### Step 3: Calculate Moles of H₂SO₄
From the balanced equation, we see that 1 mole of H₂SO₄ reacts with 2 moles of NaOH. Therefore, the moles of H₂SO₄ can be calculated as:

**Moles of H₂SO₄:**
\[ \text{Moles of H}_2\text{SO}_4 = \frac{\text{Moles of NaOH}}{2} = \frac{0.0025}{2} = 0.00125 \, \text{moles} \]

### Step 4: Calculate Moles of H₂SO₄ in 1 L of Diluted Solution
Since 25 cm³ of the diluted solution contains 0.00125 moles of H₂SO₄, we can find the moles in 1 L (1000 cm³) of the diluted solution:

**Moles of H₂SO₄ in 1 L:**
\[ \text{Moles in 1 L} = 0.00125 \, \text{moles} \times \frac{1000 \, \text{cm}^3}{25 \, \text{cm}^3} = 0.05 \, \text{moles} \]

### Step 5: Calculate Mass of H₂SO₄
The molar mass of H₂SO₄ is approximately 98 g/mol. Therefore, the mass of H₂SO₄ in 1 L of the diluted solution is:

**Mass of H₂SO₄:**
\[ \text{Mass} = \text{Moles} \times \text{Molar Mass} = 0.05 \, \text{moles} \times 98 \, \text{g/mol} = 4.9 \, \text{g} \]

### Step 6: Calculate Mass of the Original Battery Acid Solution
The density of the battery acid is given as 1.285 g/cm³. The volume of the battery acid used is 10 cm³. Therefore, the mass of the battery acid is:

**Mass of Battery Acid:**
\[ \text{Mass} = \text{Density} \times \text{Volume} = 1.285 \, \text{g/cm}^3 \times 10 \, \text{cm}^3 = 12.85 \, \text{g} \]

### Step 7: Calculate the Percentage of H₂SO₄ by Mass
Finally, we can calculate the percentage of sulfuric acid by mass in the battery acid:

**Percentage of H₂SO₄:**
\[ \text{Percentage} = \left( \frac{\text{Mass of H}_2\text{SO}_4}{\text{Mass of Battery Acid}} \right) \times 100 = \left( \frac{4.9 \, \text{g}}{12.85 \, \text{g}} \right) \times 100 \approx 38.2\% \]

### Final Answer
The percentage of sulfuric acid by mass in the battery acid is approximately **38.2%**.

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