# Question #4cfc0

##### 1 Answer

#### 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*, *milligrams per liter*,

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

Now, here's how you can think about *parts per million*,

In order to find a solution's ppm concentration, you must determine how many *grams* of solute you have in

#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

#"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

The problem tells you that

#"1 L " = " 1 kg"#

which means that your sodium fluoride solution contains

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

#color(green)(bar(ul(|color(white)(a/a)color(black)("ppm NaF" = "1.218 mg NaF"/("1 kg solvent") = "1.2 ppm")color(white)(a/a)|)))#

The answer is rounded to two **sig figs**.