# How do I calculate molality to percent by mass?

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

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

##### 1 Answer

#### Answer:

#### Explanation:

Your goal here is to figure out the number of grams of solute present for every **percent concentration by mass**,

Now, you know that the solution has a **molality** equal to **moles** of rubidium nitrate, the solute, **for every** **solvent**.

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

#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 nitratefor every#"100 g"# of the solution.

The answer is rounded to three **sig figs**.