The amount of `H_2SO_4` present in 400 mL of 0.1 M solution of the acid is -

Options

A. 2.45 g
B. 3.92 g
C. 9.80 g
D. 4.9 g

Answer

To find the amount of \( H_2SO_4 \) present in 400 mL of a 0.1 M solution, we can follow these steps:

### Step-by-Step Solution:

1. **Understand Molarity**: Molarity (M) is defined as the number of moles of solute per liter of solution. The formula for molarity is:
\[ M = \frac{\text{Number of moles of solute}}{\text{Volume of solution in liters}} \]

2. **Convert Volume to Liters**: The volume of the solution is given as 400 mL. To convert this to liters:
\[ \text{Volume in liters} = \frac{400 \text{ mL}}{1000} = 0.4 \text{ L} \]

3. **Calculate Number of Moles**: Rearranging the molarity formula to find the number of moles of solute:
\[ \text{Number of moles} = M \times \text{Volume in liters} \]
Substituting the values:
\[ \text{Number of moles} = 0.1 \, \text{M} \times 0.4 \, \text{L} = 0.04 \, \text{moles} \]

4. **Find Molecular Mass of \( H_2SO_4 \)**: The molecular mass of sulfuric acid \( H_2SO_4 \) can be calculated as follows:
- Hydrogen (H): 2 atoms × 1 g/mol = 2 g/mol
- Sulfur (S): 1 atom × 32 g/mol = 32 g/mol
- Oxygen (O): 4 atoms × 16 g/mol = 64 g/mol
- Total molecular mass = 2 + 32 + 64 = 98 g/mol

5. **Calculate the Mass of \( H_2SO_4 \)**: Now, we can find the mass of \( H_2SO_4 \) using the number of moles and the molecular mass:
\[ \text{Mass} = \text{Number of moles} \times \text{Molecular mass} \]
Substituting the values:
\[ \text{Mass} = 0.04 \, \text{moles} \times 98 \, \text{g/mol} = 3.92 \, \text{g} \]

### Final Answer:
The amount of \( H_2SO_4 \) present in 400 mL of 0.1 M solution is **3.92 grams**.

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