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

1 Answer
Apr 30, 2017

Answer:

Here's how you can do that.

Explanation:

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 water, the solvent.