# How many grams of "KNO"_3 should be dissolved in water to make "500.0 g" of a "20.0-ppm" solution?

## 1) 1.00 x 10^-3 g 2) 1.00 x 10^-4 g 3) 1.00 x 10^-1 g 4) 1.00 x 10^-2 g

Feb 28, 2018

$1.00 \cdot {10}^{- 2}$ $\text{g}$

#### Explanation:

As you know, a solution's concentration in parts per million tells you the number of grams of solute present for every

${10}^{6} = 1 , 000 , 000$

grams of the solution. In your case, the target solution must have a concentration of $\text{20.0 ppm}$, which implies that it must contain $\text{20.0 g}$ of potassium nitrate, the solute, for every ${10}^{6}$ $\text{g}$ of the solution.

This means that the mass of potassium nitrate needed to make this solution is equal to

$500.0 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{g solution"))) * "20.0 g KNO"_3/(10^6color(red)(cancel(color(black)("g solution")))) = color(darkgreen)(ul(color(black)(1.00 * 10^(-2) quad "g}}}}$

The answer is rounded to three sig figs.

So, if you dissolve $1.00 \cdot {10}^{- 2}$ $\text{g}$ of potassium nitrate in enough water to get a total mass of the solution to $\text{500.0 g}$, you will have a $\text{20.0-ppm}$ potassium nitrate solution.