# How do I calculate molality to percent by mass?

## Convert 2.35 mol/kg RbNO3 (aq) solution to percent by mass.

Feb 21, 2018

25.7%

#### Explanation:

Your goal here is to figure out the number of grams of solute present for every $\text{100 g}$ of the solution, i.e. the solution's percent concentration by mass, $\text{% m/m}$.

Now, you know that the solution has a molality equal to ${\text{2.35 mol kg}}^{- 1}$. This tells you that this solution contains $2.35$ moles of rubidium nitrate, the solute, for every $\text{1 kg}$ of water, the solvent.

To make the calculations easier, pick a sample of this solution that contains exactly $\text{1 kg" = 10^3 quad "g}$ of water. As we've said, this sample will also contain $2.35$ moles of rubidium nitrate.

Use the molar mass of the solute to convert the number of moles to grams

2.35 color(red)(cancel(color(black)("moles RbNO"_3))) * "147.473 g"/(1color(red)(cancel(color(black)("mole RbNO"_3)))) = "346.56 g"

This means that the total mass of the sample is equal to

$\text{346.56 g " + quad 10^3 quad "g" = "1346.56 g}$

So, you know that you have $\text{346.56 g}$ of rubidium nitrate in $\text{1346.56 g}$ of the solution, so you can say that $\text{100 g}$ of this solution will contain

100 color(red)(cancel(color(black)("g solution"))) * "346.56 g RbNO"_3/(1346.56 color(red)(cancel(color(black)("g solution")))) = "25.7 g RbNO"_3

This means that the solution's percent concentration by mass is equal to

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{{\text{% m/m" = "25.7% RbNO}}_{3}}}}$

This tells you that you get $\text{25.7 g}$ of rubidium nitrate for every $\text{100 g}$ of the solution.

The answer is rounded to three sig figs.