# Question #629e9

##### 1 Answer

#### Explanation:

A solution's **osmotic concentration** is a measure of how many *osmoles of particles of solute* it contains **per liter**.

As you know, an **osmole** is defined as **one mole** of particles of solute that contribute to a solution's *osmotic pressure*.

In the case of a **non-electrolyte** such as glucose, the **osmolarity** of the solution will actually be **equal** to the **molarity** of the solution.

That happens because non-electrolytes **do not** dissociate in aqueous solution. This means that **every mole** of glucose added to the solution will **not** dissociate to produce *osmoles of particles of solute*.

Now, let's pick a sample of this solution to work with. To make calculations easier, let's pick a

As you know, a **by mass** solution of glucose will contain **for every** **of solution**. This means that our sample will contain exactly

To convert this to *moles* of glucose, sue the compound's **molar mass**

#0.5 color(red)(cancel(color(black)("g"))) * "1 mole glucose"/(180.2color(red)(cancel(color(black)("g")))) = "0.002775 moles glucose"#

Since every mole will produce one osmole of particles of solute, you will have

#0.002775 color(red)(cancel(color(black)("moles"))) * "1 osmole"/(1color(red)(cancel(color(black)("mole")))) = "0.002775 osmoles"#

Now, you need to find the sample's **volume**. Since the concentration of glucose is so low, you can say that the density of the solution is approximately equal to that of water, which in turn is approximately equal to

You will have

#100color(red)(cancel(color(black)("g solution"))) * "1 mL"/(1color(red)(cancel(color(black)("g solution")))) = "100 mL"#

Remember, osmolarity is calculated **per liter of solution**, so convert the volume from *milliliters* to *liters*. You will have

#"osmolarity" = "0.002775 osmoles"/(100 * 10^(-3)"L") = color(green)(|bar(ul(color(white)(a/a)color(black)("0.028 osmol L"^(-1))color(white)(a/a)|)))#

I'll leave the answer rounded to two **sig figs**.