Question

Question #71d05

Answer 1

Here's what I got.

Explanation:

Osmolarity is simply a measure of how many osmoles of solute you get in `"1 L"` of solution.

Now, an osmole is simply `1` mole of particles that contribute to a solution's osmotic pressure. In other words, an osmole is a mole of particles produced in solution when `1` mole of solute is dissolved in water.

In order to find this solution's osmolarity, you must first find its molarity, which is defined as the number of moles of solute present in `"1 L"` of solution.

Assuming that your glucose solution is `3.8%` glucose by mass, you can say that it will contain `"3.8 g"` of glucose for every `"100 g"` of solution.

So, let's pick a `"100-g"` sample of this solution. To convert the mass of glucose to moles, use the compound's molar mass

`3.8 color(red)(cancel(color(black)("g"))) * "1 mole glucose"/(180.16color(red)(cancel(color(black)("g")))) = "0.0211 moles glucose"`

Now, in order to find the number of moles present in `"1 L"` of solution, you need to know the density of the solution. Since no information about that was given to you, and seeing how `3.8%` is still a very low concentration of glucose, you can assume that this solution has a density that is very close to that of pure water.

Let's say that the density of this solution is about `"1.01 g mL"^(-1)`. This means that your sample will have a volume of

`100 color(red)(cancel(color(black)("g solution"))) * "1 mL solution"/(1.01color(red)(cancel(color(black)("g solution")))) = "99.01 mL"`

The number of moles of glucose present in `"1 L"` of this solution will thus be equal to

`1 color(red)(cancel(color(black)("L solution"))) * (10^3color(red)(cancel(color(black)("mL"))))/(1color(red)(cancel(color(black)("L")))) * "0.0211 moles glucose"/(99.01 color(red)(cancel(color(black)("mL solution")))) = "0.213 moles glucose"`

Since this is the number of moles of glucose present in `"1 L"` of solution, you can say that the molarity of the solution will be equal to

`"molarity = 0.213 mol L"^(-1)`

Now, glucose is a non-electrolyte, which means that it does not dissociate when dissolved in water.

Consequently, every mole of glucose dissolved in water will create `1` mole of particles of solute, i.e. `1` osmole.

`"1 mole glucose = 1 osmole"`

Therefore, the osmolarity of this solution will be

`(0.213 color(red)(cancel(color(black)("moles glucose"))))/"1 L solution" * "1 osmole"/(1color(red)(cancel(color(black)("mole glucose")))) = color(darkgreen)(ul(color(black)("0.213 osmol L"^(-1))))`

I'll leave the answer rounded to three sig figs, but keep in mind that you only have one significant figure for the percent concentration of the solution.