Question #d0c11

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.