# Question 3917f

Apr 10, 2017

$\text{230 g KI}$

#### Explanation:

For starters, you know that we use molarity as a measure of the number of moles of solute present for every $\text{1 L} = {10}^{3}$ $\text{mL}$ of solution.

This means that a $\text{2.8-M}$ potassium iodide solution will contain $2.8$ moles of potassium iodide, the solute, for every ${10}^{3}$ $\text{mL}$ of solution.

You can use the molarity of a solution as a conversion factor to help you convert between the number of moles of solute and the volume of the solution.

5.00 * 10^2 color(red)(cancel(color(black)("mL solution"))) * overbrace("2.8 moles KI"/(10^3color(red)(cancel(color(black)("mL solution")))))^(color(blue)(= "2.8 M KI solution")) = "1.4 moles KI"#
$1.4 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{moles KI"))) * "166.0 g"/(1color(red)(cancel(color(black)("mole KI")))) = color(darkgreen)(ul(color(black)("230 g}}}}$