Describe how you would prepare 500.0ml of an aqueous solution that is 30.0% isopropyl alcohol by volume

The solution can be prepared by adding 150.0 mL of isopropyl alcohol to 350.0 mL of water to make 30.0% isopropyl alcohol aqueous solution.

Explanation:

∵ The volume% of a solution = [(volume of the solute(isopropy)/(volume of the solution)] x 100.

The volume % = 30.0%, volume of the solution = 500.0 mL.

∴ Volume of the solute = (the volume% of a solution)(volume of the solution)/100 = (30.0%)(500.0 mL)/100 = 150.0 mL.

∵ The volume of solution = volume of the solute + volume of water

∴ Volume of water = volume of the solution - volume of the solute = (500.0 mL) - (350.0 mL) = 150.0 mL.

So, the solution can be prepared by adding 150.0 mL of isopropyl alcohol to 350.0 mL of water to make 30.0% isopropyl alcohol aqueous solution.

Molarity is moles of solute over liters of solution. idk if that is what you are talking about, but you can convert moles of solute to grams of solute using dimensional analyzes by using molar mass found on the periodic table.

Ithink iit might be