Question #9c6af

1 Answer
Aug 13, 2017

Answer:

#"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.