Question

Question #3917f

Answer 1

`"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 `"1 L" = 10^3` `"mL"` of solution.

This means that a `"2.8-M"` potassium iodide solution will contain `2.8` moles of potassium iodide, the solute, for every `10^3` `"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.

So, your solution will need

`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"`

Now all you have to do is to convert the number of moles of potassium iodide to grams by using the compound's molar mass

`1.4 color(red)(cancel(color(black)("moles KI"))) * "166.0 g"/(1color(red)(cancel(color(black)("mole KI")))) = color(darkgreen)(ul(color(black)("230 g")))`

The answer is rounded to two sig figs, the number of sig figs you have for the molarity of the solution.