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

Apr 11, 2018

$\text{137 g}$

#### Explanation:

For starters, you know that the molarity of the solution tells you how many moles of solute are present for every $\text{1.00 L}$ of this solution.

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

Since

$\text{1 L" = 10^3 quad "mL}$

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.

$0.825 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{moles KI"))) * "166.003 g"/(1color(red)(cancel(color(black)("mole KI")))) = color(darkgreen)(ul(color(black)("137 g}}}}$

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