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 amount of charge in coulombs used up by the battery is nearly

Options

A. 954180
B. 477090
C. 95418
D. 47709

Explanation

To find the amount of charge in coulombs used up by the lead storage battery, we can follow these steps:

### Step 1: Calculate the mass of `H₂SO₄` in the battery
Given:
- Volume of `H₂SO₄` solution = 3.5 L
- Density of `H₂SO₄` = 1.294 g/mL

First, convert the volume from liters to milliliters:
\[ 3.5 \, \text{L} = 3.5 \times 1000 \, \text{mL} = 3500 \, \text{mL} \]

Now, calculate the mass of the solution:
\[ \text{Mass} = \text{Volume} \times \text{Density} = 3500 \, \text{mL} \times 1.294 \, \text{g/mL} = 4529 \, \text{g} \]

### Step 2: Calculate the mass of `H₂SO₄` in the solution
The solution is 39% `H₂SO₄` by mass:
\[ \text{Mass of } H₂SO₄ = 0.39 \times 4529 \, \text{g} = 1766.31 \, \text{g} \]

### Step 3: Calculate the number of moles of `H₂SO₄`
The molar mass of `H₂SO₄` is approximately 98 g/mol:
\[ \text{Moles of } H₂SO₄ = \frac{\text{Mass}}{\text{Molar Mass}} = \frac{1766.31 \, \text{g}}{98 \, \text{g/mol}} \approx 18.00 \, \text{moles} \]

### Step 4: Determine the change in concentration after discharge
After discharge, the density of `H₂SO₄` falls to 1.139 g/mL, which corresponds to a 20% solution by mass.

Calculate the new mass of the solution:
\[ \text{New Mass} = 3.5 \, \text{L} \times 1000 \, \text{mL/L} \times 1.139 \, \text{g/mL} = 3986.5 \, \text{g} \]

Now, calculate the mass of `H₂SO₄` in the new solution:
\[ \text{Mass of } H₂SO₄ = 0.20 \times 3986.5 \, \text{g} = 797.3 \, \text{g} \]

### Step 5: Calculate the moles of `H₂SO₄` after discharge
\[ \text{Moles of } H₂SO₄ = \frac{797.3 \, \text{g}}{98 \, \text{g/mol}} \approx 8.13 \, \text{moles} \]

### Step 6: Calculate the moles of `H₂SO₄` consumed
\[ \text{Moles consumed} = 18.00 \, \text{moles (initial)} - 8.13 \, \text{moles (final)} \approx 9.87 \, \text{moles} \]

### Step 7: Calculate the charge used
From the reduction reaction, we know that 2 moles of `H₂SO₄` require 2 Faraday (or 2 × 96500 C) of charge. Therefore, the charge for 9.87 moles of `H₂SO₄` is:
\[ \text{Charge} = \left( \frac{2 \, \text{Faraday}}{2 \, \text{moles}} \right) \times 96500 \, \text{C} \times 9.87 \, \text{moles} \]
\[ \text{Charge} = 96500 \, \text{C} \times 9.87 \approx 952,155 \, \text{C} \]

### Final Answer:
The amount of charge used up by the battery is approximately **952,155 C**.

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