# Question 4cfc0

Nov 7, 2016

$\text{1.2 ppm}$

#### Explanation:

The first thing you need to do here is to use the molar mass of sodium fluoride to convert the concentration from moles per liter, ${\text{mol L}}^{- 1}$, to milligrams per liter, ${\text{mg L}}^{- 1}$.

You will have

2.9 * 10^(-5)color(red)(cancel(color(black)("mol")))/"L" * (42.0 color(red)(cancel(color(black)("g"))))/(1color(red)(cancel(color(black)("mole NaF")))) * (10^3"mg")/(1color(red)(cancel(color(black)("g")))) = "1.218 mg L"^(-1)

You can thus say that $\text{1.00 L}$ of this solution contains $\text{1.218 mg}$ of sodium fluoride.

Now, here's how you can think about parts per million, $\text{ppm}$, concentrations.

In order to find a solution's ppm concentration, you must determine how many grams of solute you have in ${10}^{6} \text{g}$ of solvent.

color(blue)(bar(ul(|color(white)(a/a)color(black)("ppm" = "grams of solute"/(10^6"grams of solvent"))color(white)(a/a)|))) -> use this as a starting point

You can rewrite this using milligrams of solute and kilograms of solvent

(color(red)(cancel(color(black)("g")))"solute")/(10^6"g solvent") * (10^3"mg")/(1color(red)(cancel(color(black)("g")))) = (10^3"mg solute")/(10^6"g solvent") = "1 mg solute"/(10^3"g solvent")#

This is then equivalent to

$\text{1 mg solute"/(color(blue)(cancel(color(black)(10^3))) color(red)(cancel(color(black)("g")))"solvent") * (color(blue)(cancel(color(black)(10^3)))color(red)(cancel(color(black)("g"))))/("1 kg") = "1 mg solute"/"1 kg solvent}$

Therefore, you can say that a solution that contains $\text{1 mg}$ of solute in ${10}^{3} \text{g" = "1 kg}$ of solvent will have a concentration of $\text{1 ppm}$.

The problem tells you that

$\text{1 L " = " 1 kg}$

which means that your sodium fluoride solution contains $\text{1.218 mg}$ of sodium fluoride, the solute, in $\text{1 kg}$ of water, the solvent.

Therefore, you can say that the ppm concentration is equal to

$\textcolor{g r e e n}{\overline{\underline{| \textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{\text{ppm NaF" = "1.218 mg NaF"/("1 kg solvent") = "1.2 ppm}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

The answer is rounded to two sig figs.