Question

Question #9c6af

Answer 1

`"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 `"1 kg" = 10^3` `"g"` of solvent.

In your case, you know that this solution is `"16% m/m"` urea, which implies that you get `"16 g"` of urea, the solute, for every `"100 g"` of solution.

If you take a `"100-g"` sample of this solution, you can say that it contains `"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 `"1 kg" = 10^3` `"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 `10^3color(white)(.)"g" = "1 kg"` of solvent, which means that its molality will be equal to

`color(darkgreen)(ul(color(black)("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.