A lead storage battery consists of a lead anode and a grid of lead packed with lead dioxide as the cathode. The electrolyte taken is 39% `H_(2)SO_(4)` by mass having a density of 1.294 g `mL^(-1)`. The battery holds 3.5 L of the acid. During the discharge of the battery, the density `H_(2)SO_(4)` falls from 1.294 g `mL^(-1)` to 1.139 g `mL^(-1)` which is 20% `H_(2)SO_(4)` by mass
Q. The number of ampere-hour for which the battery must have been used is

Options

A. 2650.5
B. 265.05
C. 26.505
D. 2.6505

Explanation

To solve the problem step by step, we need to determine the amount of sulfuric acid (H₂SO₄) consumed during the discharge of the battery and then calculate the total charge in ampere-hours (Ah) based on the amount of H₂SO₄ used.

### Step 1: Calculate the initial mass of H₂SO₄ in the battery

1. **Volume of the electrolyte**: 3.5 L = 3500 mL
2. **Density of the electrolyte**: 1.294 g/mL
3. **Mass of the electrolyte**:
\[ \text{Mass} = \text{Volume} \times \text{Density} = 3500 \, \text{mL} \times 1.294 \, \text{g/mL} = 4529 \, \text{g} \]

4. **Mass percent of H₂SO₄**: 39%
5. **Mass of H₂SO₄**:
\[ \text{Mass of H₂SO₄} = \text{Mass of electrolyte} \times \frac{39}{100} = 4529 \, \text{g} \times 0.39 = 1766.31 \, \text{g} \]

### Step 2: Calculate the final mass of H₂SO₄ after discharge

1. **Final density of the electrolyte**: 1.139 g/mL
2. **Mass of the electrolyte after discharge**:
\[ \text{Mass} = 3500 \, \text{mL} \times 1.139 \, \text{g/mL} = 3986.5 \, \text{g} \]

3. **Mass percent of H₂SO₄ after discharge**: 20%
4. **Mass of H₂SO₄ after discharge**:
\[ \text{Mass of H₂SO₄} = 3986.5 \, \text{g} \times \frac{20}{100} = 797.3 \, \text{g} \]

### Step 3: Calculate the mass of H₂SO₄ consumed

1. **Mass of H₂SO₄ consumed**:
\[ \text{Mass consumed} = \text{Initial mass} - \text{Final mass} = 1766.31 \, \text{g} - 797.3 \, \text{g} = 969.01 \, \text{g} \]

### Step 4: Calculate the number of moles of H₂SO₄ consumed

1. **Molar mass of H₂SO₄**: 98 g/mol
2. **Moles of H₂SO₄ consumed**:
\[ \text{Moles} = \frac{\text{Mass}}{\text{Molar mass}} = \frac{969.01 \, \text{g}}{98 \, \text{g/mol}} \approx 9.88 \, \text{mol} \]

### Step 5: Calculate the total charge (in Coulombs)

1. **Faraday's constant**: 96500 C/mol
2. **Total charge**:
\[ \text{Charge} = \text{Moles} \times \text{Faraday's constant} = 9.88 \, \text{mol} \times 96500 \, \text{C/mol} \approx 9533400 \, \text{C} \]

### Step 6: Convert charge to ampere-hours

1. **Convert Coulombs to Ampere-hours**:
\[ 1 \, \text{Ah} = 3600 \, \text{C} \]
2. **Total charge in Ah**:
\[ \text{Ampere-hours} = \frac{9533400 \, \text{C}}{3600 \, \text{C/Ah}} \approx 2642.25 \, \text{Ah} \]

### Final Answer
The number of ampere-hours for which the battery must have been used is approximately **2642.25 Ah**.

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