Battery acid is 4.27 `MH_(2)SO_(4)` and has density of 1.25 g `mL^(-1)`. What is the molality of `H_(2)SO_(4)` in this solution?

Explanation

To find the molality of \( H_2SO_4 \) in the battery acid solution, we will follow these steps:

### Step 1: Calculate the weight of \( H_2SO_4 \) in 1 liter of solution.
Given that the molarity (M) of the solution is 4.27 M, we can use the formula:

\[ \text{Weight of } H_2SO_4 = \text{Molarity} \times \text{Molar Mass} \times \text{Volume} \]

The molar mass of \( H_2SO_4 \) is calculated as follows:
- Atomic mass of Hydrogen (H) = 1 g/mol
- Atomic mass of Sulfur (S) = 32 g/mol
- Atomic mass of Oxygen (O) = 16 g/mol (and there are 4 O atoms)

So, the molar mass of \( H_2SO_4 \) is:

\[ Molar\ mass\ of\ H_2SO_4 = (2 \times 1) + (1 \times 32) + (4 \times 16) = 2 + 32 + 64 = 98\ g/mol \]

Now, substituting the values into the weight calculation:

\[ \text{Weight of } H_2SO_4 = 4.27\ mol/L \times 98\ g/mol \times 1\ L = 418.46\ g \]

### Step 2: Calculate the weight of the solution.
The density of the solution is given as 1.25 g/mL. Therefore, the weight of 1 liter (1000 mL) of the solution is:

\[ \text{Weight of solution} = \text{Density} \times \text{Volume} = 1.25\ g/mL \times 1000\ mL = 1250\ g \]

### Step 3: Calculate the weight of the solvent (water).
To find the weight of the solvent, we subtract the weight of \( H_2SO_4 \) from the total weight of the solution:

\[ \text{Weight of solvent} = \text{Weight of solution} - \text{Weight of } H_2SO_4 \]

\[ \text{Weight of solvent} = 1250\ g - 418.46\ g = 831.54\ g \]

### Step 4: Calculate the molality of \( H_2SO_4 \).
Molality (m) is defined as the number of moles of solute per kilogram of solvent. First, we need to convert the weight of the solvent from grams to kilograms:

\[ \text{Weight of solvent in kg} = \frac{831.54\ g}{1000} = 0.83154\ kg \]

Now, we can calculate the molality:

\[ \text{Number of moles of } H_2SO_4 = \frac{\text{Weight of } H_2SO_4}{\text{Molar mass of } H_2SO_4} = \frac{418.46\ g}{98\ g/mol} \approx 4.27\ mol \]

Now, substituting the values into the molality formula:

\[ \text{Molality} = \frac{\text{Number of moles of solute}}{\text{Weight of solvent in kg}} = \frac{4.27\ mol}{0.83154\ kg} \approx 5.14\ mol/kg \]

### Final Answer:
The molality of \( H_2SO_4 \) in the solution is approximately **5.14 mol/kg**.

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