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

Jun 29, 2017

$\text{13 g NaF}$

#### Explanation:

As you know, the molality of a solution represents the number of moles of solute present for every $\text{1 kg}$ 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 $\text{1 kg}$ of solvent.

In your case, a $\text{0.400-m}$ sodium fluoride solution will contain $0.400$ moles of sodium fluoride, the solute, for every $\text{1 kg}$ of water, the solvent.

As you know, you have

$\text{1 kg} = {10}^{3}$ $\text{g}$

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 $\text{1 kg}$ 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

$0.300 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{moles NaF"))) * "41.988 g"/(1color(red)(cancel(color(black)("mole NaF")))) = color(darkgreen)(ul(color(black)("13 g}}}}$

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