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.