Question

Question #d0c11

Answer 1

Here's how you can do that.

Explanation:

You can determine the amount of solute dissolved to make `"500 mL"` of `"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 `"1 L" = 10^3` `"mL"` of solution.

In your case, a `"0.2-M"` potassium iodide solution will contain `0.2` moles of potassium iodide, the solute, for every `10^3` `"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 color(red)(cancel(color(black)("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.