# Question d0c11

Aug 9, 2017

Here's how you can do that.

#### Explanation:

You can determine the amount of solute dissolved to make $\text{500 mL}$ of $\text{0.2 M}$ potassium iodide solution by using the fact that a solution's molarity is a measure of the number of moles of solute present for every $\text{1 L} = {10}^{3}$ $\text{mL}$ of solution.

In your case, a $\text{0.2-M}$ potassium iodide solution will contain $0.2$ moles of potassium iodide, the solute, for every ${10}^{3}$ $\text{mL}$ of solution.

You can use this information to find the number of moles of potassium iodide present in your sample

500 color(red)(cancel(color(black)("mL solution"))) * overbrace("0.2 moles KI"/(10^3color(red)(cancel(color(black)("mL solution")))))^(color(blue)("= 0.2 M KI solution")) = "0.1 moles KI"#

Now, to convert the number of moles to grams, use the molar mass of potassium iodide

$0.1 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{moles KI"))) * "166.0 g"/(1color(red)(cancel(color(black)("mole KI")))) = color(darkgreen)(ul(color(black)("20 g KI}}}}$

The answer must be rounded to one significant figure, the number of sig figs you have for the volume and molarity of the solution.