If `100 mL` of `H_(2) SO_(4)` and `100 mL` of `H_(2)O` are mixed, the mass percent of `H_(2) SO_(4)` in the resulting solution `(d_(H_(2)SO_(4)) = 0.09 g mL^(-1), d_(H_(2)O) = 1.0 mL^(-1))`

Options

A. 90
B. 47.36
C. 50
D. 60

Explanation

To find the mass percent of \( H_2SO_4 \) in the resulting solution when \( 100 \, \text{mL} \) of \( H_2SO_4 \) is mixed with \( 100 \, \text{mL} \) of \( H_2O \), we can follow these steps:

### Step 1: Calculate the mass of \( H_2SO_4 \)

We know that mass can be calculated using the formula:
\[ \text{Mass} = \text{Density} \times \text{Volume} \]
Given:
- Density of \( H_2SO_4 \) (\( d_{H_2SO_4} \)) = \( 0.09 \, \text{g/mL} \)
- Volume of \( H_2SO_4 \) = \( 100 \, \text{mL} \)

Now, substituting the values:
\[ \text{Mass of } H_2SO_4 = 0.09 \, \text{g/mL} \times 100 \, \text{mL} = 9 \, \text{g} \]

### Step 2: Calculate the mass of \( H_2O \)

Using the same formula for water:
- Density of \( H_2O \) (\( d_{H_2O} \)) = \( 1.0 \, \text{g/mL} \)
- Volume of \( H_2O \) = \( 100 \, \text{mL} \)

Now, substituting the values:
\[ \text{Mass of } H_2O = 1.0 \, \text{g/mL} \times 100 \, \text{mL} = 100 \, \text{g} \]

### Step 3: Calculate the total mass of the solution

The total mass of the solution is the sum of the mass of \( H_2SO_4 \) and the mass of \( H_2O \):
\[ \text{Mass of solution} = \text{Mass of } H_2SO_4 + \text{Mass of } H_2O = 9 \, \text{g} + 100 \, \text{g} = 109 \, \text{g} \]

### Step 4: Calculate the mass percent of \( H_2SO_4 \)

The mass percent of \( H_2SO_4 \) in the solution can be calculated using the formula:
\[ \text{Mass percent of } H_2SO_4 = \left( \frac{\text{Mass of } H_2SO_4}{\text{Mass of solution}} \right) \times 100 \]
Substituting the values:
\[ \text{Mass percent of } H_2SO_4 = \left( \frac{9 \, \text{g}}{109 \, \text{g}} \right) \times 100 \approx 8.26\% \]

### Final Answer:
The mass percent of \( H_2SO_4 \) in the resulting solution is approximately \( 8.26\% \).

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