Let a fully charged lead storage battery contains 1.5 L 5 M `H_(2)SO_(4)`. What will be the concentration of `H_(2)SO_(4)` in the battery after 2.5 ampere current is drawn from the battery for 6 hour?

Options

A. 4.626 M
B. 0.1865 M
C. 0.373 M
D. 9.627 M

Answer

To solve the problem, we need to determine the concentration of sulfuric acid (H₂SO₄) in the battery after a certain amount of current has been drawn for a specific time. We will use Faraday's laws of electrolysis to find the number of moles of H₂SO₄ that have reacted.

### Step-by-Step Solution:

1. **Identify the Initial Conditions:**
- Volume of H₂SO₄ = 1.5 L
- Initial concentration of H₂SO₄ = 5 M

2. **Calculate the Initial Number of Moles of H₂SO₄:**
\[ \text{Number of moles} = \text{Concentration} \times \text{Volume} = 5 \, \text{mol/L} \times 1.5 \, \text{L} = 7.5 \, \text{moles} \]

3. **Determine the Total Charge Passed:**
- Current (I) = 2.5 A
- Time (t) = 6 hours = 6 × 3600 seconds = 21600 seconds
\[ \text{Total charge (Q)} = I \times t = 2.5 \, \text{A} \times 21600 \, \text{s} = 54000 \, \text{C} \]

4. **Calculate the Number of Moles of Electrons:**
- Faraday's constant (F) = 96500 C/mol
\[ \text{Number of moles of electrons} = \frac{Q}{F} = \frac{54000 \, \text{C}}{96500 \, \text{C/mol}} \approx 0.558 \, \text{moles} \]

5. **Determine the n-factor for H₂SO₄:**
- H₂SO₄ can release 2 H⁺ ions, so the n-factor = 2.

6. **Calculate the Number of Moles of H₂SO₄ Reacted:**
\[ \text{Moles of H₂SO₄ reacted} = \frac{\text{Number of moles of electrons}}{n} = \frac{0.558}{2} \approx 0.279 \, \text{moles} \]

7. **Calculate the Remaining Moles of H₂SO₄:**
\[ \text{Remaining moles of H₂SO₄} = \text{Initial moles} - \text{Moles reacted} = 7.5 - 0.279 \approx 7.221 \, \text{moles} \]

8. **Calculate the New Concentration of H₂SO₄:**
\[ \text{New concentration} = \frac{\text{Remaining moles}}{\text{Volume}} = \frac{7.221 \, \text{moles}}{1.5 \, \text{L}} \approx 4.814 \, \text{M} \]

### Final Answer:
The concentration of H₂SO₄ in the battery after drawing 2.5 A for 6 hours is approximately **4.814 M**.