Question

Question #4cfc0

Answer 1

`"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, `"mol L"^(-1)`, to milligrams per liter, `"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 `"1.00 L"` of this solution contains `"1.218 mg"` of sodium fluoride.

Now, here's how you can think about parts per million, `"ppm"`, concentrations.

In order to find a solution's ppm concentration, you must determine how many grams of solute you have in `10^6"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

`"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 `"1 mg"` of solute in `10^3"g" = "1 kg"` of solvent will have a concentration of `"1 ppm"`.

The problem tells you that

`"1 L " = " 1 kg"`

which means that your sodium fluoride solution contains `"1.218 mg"` of sodium fluoride, the solute, in `"1 kg"` of water, the solvent.

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.