Question #4cfc0

1 Answer
Nov 7, 2016

Answer:

#"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.