Calculate the molality of 90% `H_2SO_4` (weight/volume). The density of solution is `1.80 g mL^(-1)`.

Answer

To calculate the molality of a 90% H₂SO₄ (weight/volume) solution with a density of 1.80 g/mL, follow these steps:

### Step 1: Understand the Concentration
The 90% weight/volume means that there are 90 grams of H₂SO₄ in 100 mL of solution.

### Step 2: Calculate the Mass of the Solution
Using the density of the solution, we can find the mass of the solution.
Density = Mass/Volume
Rearranging gives us:
Mass = Density × Volume
Mass of solution = 1.80 g/mL × 100 mL = 180 grams

### Step 3: Calculate the Mass of the Solvent (Water)
To find the mass of the solvent (water), we subtract the mass of the solute (H₂SO₄) from the mass of the solution.
Mass of water = Mass of solution - Mass of H₂SO₄
Mass of water = 180 g - 90 g = 90 grams

### Step 4: Convert Mass of Solvent to Kilograms
Since molality is defined as moles of solute per kilogram of solvent, we need to convert the mass of water from grams to kilograms.
Mass of water in kg = 90 g × (1 kg / 1000 g) = 0.090 kg

### Step 5: Calculate the Number of Moles of H₂SO₄
To find the number of moles of H₂SO₄, we use the molar mass of H₂SO₄. The molar mass of H₂SO₄ is approximately 98 g/mol.
Number of moles = Mass / Molar mass
Number of moles of H₂SO₄ = 90 g / 98 g/mol ≈ 0.9184 moles

### Step 6: Calculate the Molality
Now we can calculate the molality using the formula:
Molality (m) = Number of moles of solute / Mass of solvent in kg
Molality = 0.9184 moles / 0.090 kg ≈ 10.204 m

### Final Answer
The molality of the 90% H₂SO₄ solution is approximately **10.204 m**.

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