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

##### 1 Answer

#### Answer:

#### Explanation:

For starters, you know that the molarity of the solution tells you how many moles of solute are present for every

More specifically, you know that in order for the solution to have a molarity of **moles** of potassium iodide, the solute, for every

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.