Find the weight of `H_(2) SO_(4)` in `1200 mL` of a solution of `0.2 N` strength.

Explanation

To find the weight of \( H_2SO_4 \) in 1200 mL of a solution with a normality of 0.2 N, we can follow these steps:

### Step 1: Understand Normality
Normality (N) is defined as the number of gram equivalents of solute per liter of solution. The formula for normality is given by:

\[ N = \frac{\text{Weight of solute (g)}}{\text{Equivalent weight of solute (g/equiv)} \times \text{Volume of solution (L)}} \]

### Step 2: Convert Volume from mL to L
We need to convert the volume of the solution from milliliters to liters:

\[ 1200 \, \text{mL} = \frac{1200}{1000} = 1.2 \, \text{L} \]

### Step 3: Determine the Equivalent Weight of \( H_2SO_4 \)
The equivalent weight of \( H_2SO_4 \) can be calculated using its molar mass and the number of acidic hydrogens (N). The molar mass of \( H_2SO_4 \) is calculated as follows:

- Hydrogen (H): 1 g/mol × 2 = 2 g/mol
- Sulfur (S): 32 g/mol × 1 = 32 g/mol
- Oxygen (O): 16 g/mol × 4 = 64 g/mol

Adding these together gives:

\[ \text{Molar mass of } H_2SO_4 = 2 + 32 + 64 = 98 \, \text{g/mol} \]

Since \( H_2SO_4 \) has 2 acidic hydrogens, the equivalent weight is:

\[ \text{Equivalent weight} = \frac{\text{Molar mass}}{N} = \frac{98 \, \text{g/mol}}{2} = 49 \, \text{g/equiv} \]

### Step 4: Calculate the Weight of \( H_2SO_4 \)
Using the normality formula, we can rearrange it to find the weight of the solute:

\[ \text{Weight} = N \times \text{Equivalent weight} \times \text{Volume (L)} \]

Substituting the known values:

\[ \text{Weight} = 0.2 \, \text{N} \times 49 \, \text{g/equiv} \times 1.2 \, \text{L} \]

Calculating this gives:

\[ \text{Weight} = 0.2 \times 49 \times 1.2 = 11.76 \, \text{g} \]

### Final Answer
The weight of \( H_2SO_4 \) in 1200 mL of a 0.2 N solution is **11.76 grams**.

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