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

Apr 30, 2017

Here's how you can do that.

#### Explanation:

For starters, you should know that a solution's percent concentration by mass. $\text{m/m %}$, is used to denote the number of grams of solute present for every $\text{100 g}$ of solution.

In your case, a $\text{5.00% m/m}$ solution will contain $\text{5.00 g}$ of potassium iodide, the solute, for every $\text{100 mL}$ of solution.

Now, you know that the solution must have a total mass of $\text{72.5 g}$. 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 $\text{3.63 g}$ of potassium iodide in

$\text{72.5 g " - " 3.63 g" = "68.87 g}$

of water, the solvent.