# Question b22b0

Oct 8, 2017

Here's what I got.

#### Explanation:

Start by picking a sample of this solution that contains exactly

$\text{1 kg" = 10^3color(white)(.)"g}$

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

In your case, a $\text{2-molal}$ sodium hydroxide solution will contain $2$ moles of sodium hydroxide, the solute, for every $\text{1 kg}$ of water, the solvent.

Since we picked a sample that contains $\text{1 kg}$ of water, you can say that this sample will also contain $2$ moles of sodium hydroxide.

Use the molar mass of sodium hydroxide to convert the number of mols to grams

2 color(red)(cancel(color(black)("moles NaOH"))) * "40.0 g"/(1color(red)(cancel(color(black)("mole NaOH")))) = "80.0 g"

This means that the total mass of the solution will be equal to

overbrace("80.0 g")^(color(blue)("mass of NaOH")) + overbrace(10^3color(white)(.)"g")^(color(blue)("mass of water")) = overbrace("1080 g")^(color(blue)("mass of solution"))

Now, in order to find the solution's mass by mass percent concentration, $\text{m/m %}$, you need to figure out the mass of solute present in exactly $\text{100 g}$ of the solution.

To do that, use the known composition of the sample

100 color(red)(cancel(color(black)("g solution"))) * "80.0 g NaOH"/(1080color(red)(cancel(color(black)("g solution")))) = "7.41 g NaOH"

Since this represents the mass of sodium hydroxide present in '100 g"# of the solution, you can say that the solution's mass by mass percent concentration will be

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{\text{% m/m = 7.4% NaOH}}}}$

I'll leave the answer rounded to two sig figs, but do not forget that you have one significant figure for the molality of the solution.

So the answer should be given as

$\text{% m/m = 7% NaOH}$