Question #9d8f3

The idea here is that the solution's mass by volume percent concentration, `"m/v%"`, will help you determine how many grams of glucose you have in your solution.

You know that a solution's mass by volume percent concentration tells you the number of grams of solute present for every `"100 mL"` of solution.

In your case, a `"9% m/v"` glucose solution will contain `"9 g"` of glucose, the solute, for every `"100 mL"` of solution.

This means that your sample

`"500 mL" = color(red)(5) * "100 mL"`

will contain `color(red)(5)` times more grams of glucose than what you have in `"100 mL"` of solution, i.e. than the solution's percent concentration.

Therefore, you will have

`color(red)(5) * overbrace("9 g glucose")^(color(blue)("present in 100 mL of solution")) = "45 g glucose"`

To convert this to moles, use the compound's molar mass

`45 color(red)(cancel(color(black)("g"))) * "1 mole glucose"/(180color(red)(cancel(color(black)("g")))) = "0.25 moles glucose"`

Finally, to convert this to number of molecules, use Avogadro;s constant

`0.25 color(red)(cancel(color(black)("moles glucose"))) * (6.022 * 10^(23)color(white)(.)"molecules glucose")/(1color(red)(cancel(color(black)("mole glucose"))))`

`= color(darkgreen)(ul(color(black)(2 * 10^(23)color(white)(.)"molecules glucose")))`

The answer is rounded to one significant figure, the number of sig figs you have for the volume of the solution.