Question #629e9

1 Answer
May 29, 2016


#"0.028 osmol L"^(-1)#


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 #"100-g"# sample.

As you know, a #0.5%# by mass solution of glucose will contain #"0.5 g"# of glucose for every #"100 g"# of solution. This means that our sample will contain exactly #"0.5 g"# of glucose.

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 #"1 g mL"^(-1)#.

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.