# If the mole fraction of "NaCl" in a solution is 0.125, what is the percent by mass of "NaCl"?

Jul 14, 2018

$\text{31.7% NaCl}$

#### Explanation:

The trick here is to pick a sample of this solution and use the mole fraction of sodium chloride to find the total mass of the sample and the mass of the solute.

Assuming that your solution contains sodium chloride as the solute and water as the solvent, you can say that

${\chi}_{\text{NaCl") + chi_ ("H"_ 2"O") = 1" " " "color(darkorange)("(*)}}$

Here

• ${\chi}_{\text{NaCl}}$ is the mole fraction of sodium chloride
• ${\chi}_{\text{H"_ 2"O}}$ is the mole faction of water

Now, let's assume that you pick a sample of this solution that contains exactly $1$ mole of particles, i.e. that the total number of moles present in the sample is equal to $1$.

By definition, the mole fraction of sodium chloride is equal to the number of moles of sodium chloride divided by the total number of moles present in the solution.

chi_( "NaCl") = ("moles of NaCl")/"total moles"

So in this case, you'd have

0.125 = ("moles of NaCl")/"1 mole"

which implies that the sample contains $0.125$ moles of sodium chloride. Following the same logic or using equation $\textcolor{\mathrm{da} r k \mathmr{and} a n \ge}{\text{(*)}}$, you can determine that this sample contains

$\text{1 mole" \ - \ "0.125 moles" = "0.875 moles H"_2"O}$

Next, use the molar masses of the two compounds to convert the number of moles to grams.

0.125 color(red)(cancel(color(black)("moles NaCl"))) * "58.44 g"/(1color(red)(cancel(color(black)("mole NaCl")))) = "7.3050 g"

0.875 color(red)(cancel(color(black)("moles H"_2"O"))) * "18.015 g"/(1color(red)(cancel(color(black)("mole H"_2"O")))) = "15.7631 g"

You can now say that the total mass of the sample is

$\text{7.3050 g" \ + \ "15.7631 g" = "23.0681 g}$

Finally, to find the percent by mass of sodium chloride, you need to determine the mass of sodium chloride present for every $\text{100. g}$ of this solution.

Your sample contains $\text{7.3050 g}$ of sodium chloride in $\text{23.0681 g}$ of the solution, which means that you have

100. color(red)(cancel(color(black)("g solution"))) * ("7.3050 g NaCl")/(23.0681color(red)(cancel(color(black)("g solution")))) = "31.7 g NaCl"

Therefore, you can say that the percent by mass of sodium chloride is

color(darkgreen)(ul(color(black)("% NaCl" = 31.7%)))

The answer is rounded to three sig figs, the number of sig figs you have for the mole fraction of sodium chloride.