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 find the concentration of `H₂SO₄` in the battery after a current is drawn, we can follow these steps:

### Step 1: Calculate the Initial Moles of `H₂SO₄`
Given:
- Volume of `H₂SO₄` solution = 1.5 L
- Molarity of `H₂SO₄` = 5 M

Using the formula for moles:
\[ \text{Moles of } H₂SO₄ = \text{Molarity} \times \text{Volume} = 5 \, \text{mol/L} \times 1.5 \, \text{L} = 7.5 \, \text{moles} \]

### Step 2: Calculate the Total Charge Passed
Given:
- Current (I) = 2.5 A
- Time (t) = 6 hours = 6 × 3600 seconds = 21600 seconds

Using the formula for charge:
\[ Q = I \times t = 2.5 \, \text{A} \times 21600 \, \text{s} = 54000 \, \text{C} \]

### Step 3: Calculate the Moles of Electrons Transferred
Using Faraday's constant (F = 96500 C/mol):
\[ \text{Moles of electrons} = \frac{Q}{F} = \frac{54000 \, \text{C}}{96500 \, \text{C/mol}} \approx 0.56 \, \text{moles} \]

### Step 4: Relate Moles of `H₂SO₄` to Moles of Electrons
From the cell reaction, we see that 1 mole of `H₂SO₄` corresponds to 2 moles of electrons. Therefore, the moles of `H₂SO₄` consumed can be calculated as:
\[ \text{Moles of } H₂SO₄ \text{ consumed} = \frac{0.56}{2} = 0.28 \, \text{moles} \]

### Step 5: Calculate Remaining Moles of `H₂SO₄`
Subtract the moles of `H₂SO₄` consumed from the initial moles:
\[ \text{Remaining moles of } H₂SO₄ = 7.5 \, \text{moles} - 0.28 \, \text{moles} = 7.22 \, \text{moles} \]

### Step 6: Calculate the New Concentration of `H₂SO₄`
Using the remaining moles and the original volume:
\[ \text{New concentration} = \frac{\text{Remaining moles}}{\text{Volume}} = \frac{7.22 \, \text{moles}}{1.5 \, \text{L}} \approx 4.81 \, \text{M} \]

### Final Answer:
The concentration of `H₂SO₄` in the battery after the current is drawn will be approximately **4.81 M**.

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