Question

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

Answer 1

`"137 g"`

Explanation:

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.

Since

`"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.