Question

Calculate the molality of phosphoric acid (H₃PO₄) in a 24.1% (by mass) aqueous solution. Choose the correct molality:

a) 0.110 m
b) 3.24 m
c) 0.382 m
d) 5.52 × 10⁻² m

Answer 1

The molality of phosphoric acid in a 24.1% aqueous solution is calculated by converting the mass of phosphoric acid to moles, the mass of water to kilograms, and dividing the moles of acid by the kilograms of water to get a molality of 3.24 m.

Step-by-step explanation:

To calculate the molality of phosphoric acid (H3PO4) in a 24.1% (by mass) aqueous solution, first, it is necessary to determine the mass of phosphoric acid and water. Assuming we have 100 g of the solution, 24.1 g will be phosphoric acid, and 75.9 g will be water. Next, we convert the mass of phosphoric acid to moles by dividing by its molar mass (98.00 g/mol). Then, we convert the mass of water to kilograms since molality is moles of solute per kilogram of solvent. Finally, the molality is calculated using the formula:

1. Mass of H3PO4 = 24.1 g of phosphoric acid
2. Molar mass of H3PO4 = 98.00 g/mol
3. Mass of water = 75.9 g = 0.0759 kg (convert grams to kilograms)
4. Moles of H3PO4 = Mass of H3PO4 / Molar mass of H3PO4
5. Molality (m) = Moles of H3PO4 / Mass of water (kg)

Substituting the values:

• Moles of H3PO4 = 24.1 g / 98.00 g/mol = 0.246 moles
• Molality = 0.246 moles / 0.0759 kg = 3.24 m

Therefore, the correct molality is 3.24 m.