# Question 9c6af

Aug 13, 2017

${\text{3.2 mol kg}}^{- 1}$

#### Explanation:

In order to be able to calculate the molality of the solution, you need to know the number of moles of solute present for every $\text{1 kg} = {10}^{3}$ $\text{g}$ of solvent.

In your case, you know that this solution is $\text{16% m/m}$ urea, which implies that you get $\text{16 g}$ of urea, the solute, for every $\text{100 g}$ of solution.

If you take a $\text{100-g}$ sample of this solution, you can say that it contains $\text{16 g}$ of urea and

overbrace("100 g ")^(color(blue)("mass of solution")) - overbrace("16 g")^(color(blue)("mass of solute")) = overbrace("84 g")^(color(blue)("mass of solvent"))

of water, the solvent.

You can use the composition of the sample to determine the mass of urea present in $\text{1 kg} = {10}^{3}$ $\text{g}$ of water. Keep in mind that you can do this because solutions are homogeneous mixtures, i.e. they have the same composition throughout.

10^3 color(red)(cancel(color(black)("g water"))) * "16 g urea"/(84color(red)(cancel(color(black)("g water")))) = "190.48 g urea"

To convert this to moles, use the molar mass of urea

190.48 color(red)(cancel(color(black)("g urea"))) * "1 mole urea"/(60.06color(red)(cancel(color(black)("g urea")))) = "3.17 moles urea"#

So, you know that your solution will contain $3.17$ moles of urea for every ${10}^{3} \textcolor{w h i t e}{.} \text{g" = "1 kg}$ of solvent, which means that its molality will be equal to

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{{\text{molality = 3.2 mol kg}}^{- 1}}}}$

The answer is rounded to two sig figs, the number of sig figs you have for the solution's percent concentration.