Calculate the molality of 1 litre solution of 93% `H_2SO_4` (weight/volume). The density of the solution is `1.84 g ml^(-1)`.

Answer

To calculate the molality of a 1-liter solution of 93% H₂SO₄ (weight/volume) with a density of 1.84 g/mL, we can follow these steps:

### Step 1: Determine the mass of the solution
Given that the density of the solution is 1.84 g/mL, we can calculate the mass of 1 liter (1000 mL) of the solution.

\[ \text{Mass of solution} = \text{Volume} \times \text{Density} = 1000 \, \text{mL} \times 1.84 \, \text{g/mL} = 1840 \, \text{g} \]

### Step 2: Calculate the mass of H₂SO₄ in the solution
The solution is 93% H₂SO₄ by weight/volume, which means that in 100 mL of solution, there are 93 grams of H₂SO₄. Therefore, in 1000 mL (1 liter) of solution:

\[ \text{Mass of H₂SO₄} = 93 \, \text{g} \times 10 = 930 \, \text{g} \]

### Step 3: Calculate the moles of H₂SO₄
To find the number of moles of H₂SO₄, we use its molar mass. The molar mass of H₂SO₄ is 98 g/mol.

\[ \text{Moles of H₂SO₄} = \frac{\text{Mass}}{\text{Molar Mass}} = \frac{930 \, \text{g}}{98 \, \text{g/mol}} \approx 9.49 \, \text{mol} \]

### Step 4: Calculate the mass of the solvent (water)
The mass of the solvent (water) can be found by subtracting the mass of H₂SO₄ from the total mass of the solution.

\[ \text{Mass of solvent} = \text{Mass of solution} - \text{Mass of H₂SO₄} = 1840 \, \text{g} - 930 \, \text{g} = 910 \, \text{g} \]

### Step 5: Calculate the molality
Molality (m) is defined as the number of moles of solute per kilogram of solvent. We need to convert the mass of the solvent from grams to kilograms.

\[ \text{Mass of solvent in kg} = \frac{910 \, \text{g}}{1000} = 0.910 \, \text{kg} \]

Now we can calculate the molality:

\[ \text{Molality} = \frac{\text{Moles of H₂SO₄}}{\text{Mass of solvent in kg}} = \frac{9.49 \, \text{mol}}{0.910 \, \text{kg}} \approx 10.42 \, \text{mol/kg} \]

### Final Answer
The molality of the solution is approximately **10.42 mol/kg**.

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