How many grams of sodium fluoride (NaF) are needed to prepare a 0.400 m NaF solution that contains 750 g water?

1 Answer
Jun 29, 2017

Answer:

#"13 g NaF"#

Explanation:

As you know, the molality of a solution represents the number of moles of solute present for every #"1 kg"# of solvent.

This means that in order to find a solution's molality, all you need to do is to figure out exactly how many moles of solute you have for #"1 kg"# of solvent.

In your case, a #"0.400-m"# sodium fluoride solution will contain #0.400# moles of sodium fluoride, the solute, for every #"1 kg"# of water, the solvent.

As you know, you have

#"1 kg" = 10^3# #"g"#

This means that your target solution contains

#750 color(red)(cancel(color(black)("g"))) * "1 kg"/(10^3color(red)(cancel(color(black)("g")))) = "0.750 kg" = color(blue)(3/4)* "1 kg"#

of water. Since you already know that #"1 kg"# of this solution must contain #0.400# moles of sodium fluoride, you can say that your sample will contain

#color(blue)(3/4) * overbrace("0.400 moles NaF")^(color(red)("no. of moles of NaF in 1 kg of water")) = "0.300 moles NaF"#

Finally, to convert this to grams, use the molar mass of the compound

#0.300 color(red)(cancel(color(black)("moles NaF"))) * "41.988 g"/(1color(red)(cancel(color(black)("mole NaF")))) = color(darkgreen)(ul(color(black)("13 g")))#

The answer must be rounded to two sig figs, the number of sig figs you have for the mass of water.