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. Molarity of the solution after the discharge is

Options

A. 8.136
B. 4.068
C. 2.32
D. 1.16

Explanation

To find the molarity of the sulfuric acid solution after the discharge of the lead storage battery, we can follow these steps:

### Step 1: Calculate the mass of the sulfuric acid solution before discharge.
The density of the sulfuric acid solution before discharge is given as 1.294 g/mL, and the volume of the solution is 3.5 L.

\[ \text{Mass of solution} = \text{Density} \times \text{Volume} \]
\[ \text{Mass of solution} = 1.294 \, \text{g/mL} \times 3500 \, \text{mL} = 4519 \, \text{g} \]

### Step 2: Calculate the mass of sulfuric acid (H₂SO₄) in the solution before discharge.
The solution is 39% H₂SO₄ by mass.

\[ \text{Mass of H₂SO₄} = \text{Mass of solution} \times \text{Percentage of H₂SO₄} \]
\[ \text{Mass of H₂SO₄} = 4519 \, \text{g} \times 0.39 = 1762.41 \, \text{g} \]

### Step 3: Calculate the mass of the solution after discharge.
After discharge, the density of the sulfuric acid solution falls to 1.139 g/mL.

\[ \text{Mass of solution after discharge} = \text{Density} \times \text{Volume} \]
\[ \text{Mass of solution after discharge} = 1.139 \, \text{g/mL} \times 3500 \, \text{mL} = 3986.5 \, \text{g} \]

### Step 4: Calculate the mass of sulfuric acid (H₂SO₄) in the solution after discharge.
The solution is now 20% H₂SO₄ by mass.

\[ \text{Mass of H₂SO₄ after discharge} = \text{Mass of solution after discharge} \times \text{Percentage of H₂SO₄} \]
\[ \text{Mass of H₂SO₄ after discharge} = 3986.5 \, \text{g} \times 0.20 = 797.3 \, \text{g} \]

### Step 5: Calculate the number of moles of sulfuric acid (H₂SO₄) after discharge.
The molar mass of H₂SO₄ is approximately 98 g/mol.

\[ \text{Moles of H₂SO₄} = \frac{\text{Mass of H₂SO₄}}{\text{Molar mass of H₂SO₄}} \]
\[ \text{Moles of H₂SO₄} = \frac{797.3 \, \text{g}}{98 \, \text{g/mol}} \approx 8.136 \, \text{mol} \]

### Step 6: Calculate the molarity of the solution after discharge.
Molarity (M) is defined as the number of moles of solute per liter of solution.

\[ \text{Molarity} = \frac{\text{Moles of H₂SO₄}}{\text{Volume of solution in L}} \]
\[ \text{Molarity} = \frac{8.136 \, \text{mol}}{3.5 \, \text{L}} \approx 2.32 \, \text{M} \]

### Final Answer:
The molarity of the solution after the discharge is approximately **2.32 M**.

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