Question #8c3ef

Assuming that you're working with a mass by volume percent concentration, #"m/v%"#, which tells you the number of grams of solute present for every #"100 mL"# of solution, start by calculating the mass of urea present in your sample.

#10 color(red)(cancel(color(black)("mL solution"))) * "5 g urea"/(100color(red)(cancel(color(black)("mL solution")))) = "0.5 g urea"#

Now, in order to find the solution's osmolarity, you need to find its molarity, i.e. the number of moles of solute present for every #"1 L" = 10^3# #"mL"# of solution.

To do that, calculate the number of grams of urea present in #10^3# #"mL"# of this solution

#10^3 color(red)(cancel(color(black)("mL solution"))) * "0.5 g urea"/(10color(red)(cancel(color(black)("mL solution")))) = "50 g urea"#

Now use the molar mass of urea to convert this to moles

#50 color(red)(cancel(color(black)("g"))) * "1 mole urea"/(60.06color(red)(cancel(color(black)("g")))) = "0.8325 moles urea"#

Since this represents the number of moles of urea present in #10^3# #"mL"# of solution, you can say that its molarity is equal to

#"molarity = 0.8325 mol L"^(-1)#

To find the osmolarity of the solution, which tells you the number of osmoles of solute present in #10^3# #"mL"# of solution, use the fact that urea is a non-electrolyte.

As you know, an Osmole is simply a mole of particles that contribute to a solution's osmotic pressure. In this case, the fact that urea is a non-electrolyte implies that it does not dissociate in aqueous solution to produce ions.

Consequently, you can say that #1# mole of urea is equivalent to #1# Osmole because urea is a non-electrolyte.

Therefore, the osmolarity of the solution will be equal to

#0.8325color(white)(.) color(red)(cancel(color(black)("moles")))/"1 L solution" * "1 Osmole"/(1color(red)(cancel(color(black)("mole")))) = color(darkgreen)(ul(color(black)("0.83 Osmol L"^(-1))))#

I'll leave the answer rounded to two sig figs, but keep in mind that you only have one significant figure for your values.