Question

An aqueous solution is 29.5 % H₃PO₄ by mass and the solution has a density of 1.2 g/mL. A. Using only that the solution is 29.5 % H₃PO₄, then give the masses of the solute, solvent, and solution that make the solution.

____g H₃PO₄ (solute)
____g water (solvent)
____g solution

Answer 1

To calculate the masses of the components in a 29.5% H₃PO₄ aqueous solution, assume a total mass of 100 g of solution, leading to 29.5 g of H₃PO₄ and 70.5 g of water.

Step-by-step explanation:

To solve for the masses of the solute, solvent, and solution in an aqueous solution that is 29.5% H₃PO₄ by mass with a density of 1.2 g/mL, you would use the percentages given and density to find the individual masses.

• Determine the total mass of the solution needed. For calculation purposes, assume a convenient mass of solution like 100 g, which, due to the percentage by mass, means it contains 29.5 g of H₃PO₄.
• Calculate the mass of the solute (H₃PO₄) directly from the percentage by mass. For a 100 g solution, the mass of H₃PO₄ is 29.5 g because 29.5% of 100 g is 29.5 g.
• Subtract the mass of the solute from the total mass of the solution to find the mass of the solvent (water). Therefore, the mass of water would be 100 g - 29.5 g = 70.5 g.

So, in a 100 g solution with a 29.5% concentration by mass of H₃PO₄:

• The mass of H₃PO₄ (solute) is 29.5 g
• The mass of water (solvent) is 70.5 g
• The total mass of the solution is 100 g