How do I calculate molality to percent by mass?

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

1 Answer
Feb 21, 2018

Answer:

#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.