How would you prepare 72.5 g of an aqueous solution that is 5.00% potassium iodide, KI, by mass?

Question

Owen Ambrose · Accepted Answer

Here's how you can do that.

For starters, you should know that a solution's percent concentration by mass. #"m/m %"#, is used to denote the number of grams of solute present for every #"100 g"# of solution. In your case, a #"5.00% m/m"# solution will contain #"5.00 g"# of potassium iodide, the solute, for every #"100 mL"# of solution. Now, you know that the solution must have a total mass of #"72.5 g"#. You can use the target percent concentration by mass to figure out how many grams of potassium iodide must be present in the target solution #72.5 color(red)(cancel(color(black)("g solution"))) * "5.00 g KI"/(100color(red)(cancel(color(black)("g solution")))) = "3.63 g KI"# You can thus say that the solution can be prepared by dissolving #"3.63 g"# of potassium iodide in #"72.5 g " - " 3.63 g" = "68.87 g"# of the solvent, water.

Cooper Adams · Answer

undefined To prepare 72.5 g of an aqueous solution that is 5.00% potassium iodide (KI) by mass, you would dissolve 3.63 g of potassium iodide in water to make a total solution mass of 72.5 g.