How many grams of KI are required to make 550.0 mL of a 1.50 M KI solution ?

1 Answer
Apr 11, 2018

#"137 g"#


For starters, you know that the molarity of the solution tells you how many moles of solute are present for every #"1.00 L"# of this solution.

More specifically, you know that in order for the solution to have a molarity of #"1.50 M"#, it must contain #1.50# moles of potassium iodide, the solute, for every #"1.00 L"# of the solution.


#"1 L" = 10^3 quad "mL"#

you can say that your sample must contain

#550.0 color(red)(cancel(color(black)("mL solution"))) * "1.50 moles KI"/(10^3color(red)(cancel(color(black)("mL solution")))) = "0.825 moles KI"#

To convert the number of moles of potassium iodide to grams, use the molar mass of the compound.

#0.825 color(red)(cancel(color(black)("moles KI"))) * "166.003 g"/(1color(red)(cancel(color(black)("mole KI")))) = color(darkgreen)(ul(color(black)("137 g")))#

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