# Question #8c3ef

##### 1 Answer

Here's what I got.

#### Explanation:

Assuming that you're working with a *mass by volume* percent concentration, **for every** **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

To do that, calculate the number of grams of urea present in

#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

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

To find the **osmolarity** of the solution, which tells you the number of *osmoles* of solute present in **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 **mole** of urea is equivalent to **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.