Question

How do I calculate molality to percent by mass?

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

Answer 1

`25.7%`

Explanation:

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

Now, you know that the solution has a molality equal to `"2.35 mol kg"^(-1)`. This tells you that this solution contains `2.35` moles of rubidium nitrate, the solute, for every `"1 kg"` of water, the solvent.

To make the calculations easier, pick a sample of this solution that contains exactly `"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

`"346.56 g " + quad 10^3 quad "g" = "1346.56 g"`

So, you know that you have `"346.56 g"` of rubidium nitrate in `"1346.56 g"` of the solution, so you can say that `"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

`color(darkgreen)(ul(color(black)("% m/m" = "25.7% RbNO"_3)))`

This tells you that you get `"25.7 g"` of rubidium nitrate for every `"100 g"` of the solution.

The answer is rounded to three sig figs.