Question

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

Answer 1

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.