# How many grams of KNO_3 should be used to prepare 2.00 L of a 0.500 M solution?

Mar 8, 2016

${\text{101 g KNO}}_{3}$

#### Explanation:

Molarity is simply a measure of how many moles of solute you get per liter of solution.

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \text{molarity" = "moles of solute"/"one liter of solution} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

This means that in order to find a solution's molarity, you need to know two things

• how many moles of solute it contains
• its volume expressed in liters

Now, you already know the molarity and volume of your target solution, which means that you can use the definition of molarity to find how many moles of potassium nitrate, ${\text{KNO}}_{3}$, it must contain.

Molarity is measued in moles per liter, ${\text{mol L}}^{- 1}$, or molar, $\text{M}$, which means that a $\text{0.500-M}$ solution will contain $0.500$ moles of potassium nitrate for every $\text{1 L}$ of solution.

Well, if you get "0.500 moles for very liter of solution, and your solution has a total volume of $\text{2.0 L}$, it follows that it must contain

$2 \times {\text{0.500 moles" = "1.00 moles KNO}}_{3}$

So, what mass of potassium nitrate is equivalent to one mole of the compound?

Potassium nitrate's molar mass, ${\text{101.103 g mol}}^{- 1}$, tells you the mass of one mole of potassium nitrate. In this case, the answer must be rounded to three sig figs, so

${m}_{K N {O}_{3}} = \textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} {\text{101 g KNO}}_{3} \textcolor{w h i t e}{\frac{a}{a}} |}}}$