During the discharge of a lead storage battery, the density of sulphuric acid fell from `1.294 g mL^(-1)` to `1.139 g mL^(-)`. Sulphuric acid of density `1.294 g mL^(-1)` is `39%` by weight and that of density `1.139 g mL^(-1)` is `20%` by weight. The battery hold `3.5` litre of acied and discharge. Calculate the no. of ampere hour for which the battery must have been used. The charging and discharging reactions are:
`Pb+SO_(4)^(2-) rarr PbSO_(4)+2e` (charging)
`PbO_(2)+4H^(+)+SO_(4)^(2-)+2e rarr PbSO_(4)+2H_(2)O` (discharging)

Explanation

To solve the problem regarding the lead storage battery, we need to follow these steps:

### Step 1: Calculate the mass of sulfuric acid before and after discharge.
1. **Before Discharge:**
- Density of sulfuric acid = 1.294 g/mL
- Volume of acid = 3.5 L = 3500 mL
- Mass of sulfuric acid before discharge = Density × Volume
\[ \text{Mass before} = 1.294 \, \text{g/mL} \times 3500 \, \text{mL} = 4529 \, \text{g} \]

2. **After Discharge:**
- Density of sulfuric acid = 1.139 g/mL
- Mass of sulfuric acid after discharge = Density × Volume
\[ \text{Mass after} = 1.139 \, \text{g/mL} \times 3500 \, \text{mL} = 3986.5 \, \text{g} \]

### Step 2: Calculate the mass of sulfuric acid consumed.
\[ \text{Mass consumed} = \text{Mass before} - \text{Mass after} = 4529 \, \text{g} - 3986.5 \, \text{g} = 542.5 \, \text{g} \]

### Step 3: Calculate the number of moles of sulfuric acid consumed.
- The molecular weight of sulfuric acid (H₂SO₄) = 98 g/mol.
\[ \text{Moles of H₂SO₄} = \frac{\text{Mass consumed}}{\text{Molar mass}} = \frac{542.5 \, \text{g}}{98 \, \text{g/mol}} \approx 5.53 \, \text{mol} \]

### Step 4: Determine the number of moles of electrons transferred.
- The discharging reaction shows that 1 mole of H₂SO₄ produces 2 moles of electrons.
\[ \text{Moles of electrons} = 2 \times \text{Moles of H₂SO₄} = 2 \times 5.53 \approx 11.06 \, \text{mol} \]

### Step 5: Calculate the total charge in coulombs.
- Using Faraday's constant (F = 96500 C/mol),
\[ \text{Total charge (Q)} = \text{Moles of electrons} \times F = 11.06 \, \text{mol} \times 96500 \, \text{C/mol} \approx 1066460 \, \text{C} \]

### Step 6: Convert charge to ampere-hours.
- 1 ampere-hour (Ah) = 3600 C,
\[ \text{Ampere-hours} = \frac{Q}{3600} = \frac{1066460 \, \text{C}}{3600 \, \text{C/Ah}} \approx 296.8 \, \text{Ah} \]

### Final Answer:
The battery must have been used for approximately **296.8 ampere-hours**.

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