# Question 8c3ef

Jun 14, 2017

Here's what I got.

#### Explanation:

Assuming that you're working with a mass by volume percent concentration, $\text{m/v%}$, which tells you the number of grams of solute present for every $\text{100 mL}$ 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 $\text{1 L} = {10}^{3}$ $\text{mL}$ of solution.

To do that, calculate the number of grams of urea present in ${10}^{3}$ $\text{mL}$ 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 ${10}^{3}$ $\text{mL}$ of solution, you can say that its molarity is equal to

${\text{molarity = 0.8325 mol L}}^{- 1}$

To find the osmolarity of the solution, which tells you the number of osmoles of solute present in ${10}^{3}$ $\text{mL}$ 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

$0.8325 \textcolor{w h i t e}{.} \textcolor{red}{\cancel{\textcolor{b l a c k}{{\text{moles")))/"1 L solution" * "1 Osmole"/(1color(red)(cancel(color(black)("mole")))) = color(darkgreen)(ul(color(black)("0.83 Osmol L}}^{- 1}}}}$

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