# Question #9d8f3

##### 1 Answer

#### Explanation:

The idea here is that the solution's **mass by volume percent concentration**, *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**

In your case, a

This means that your sample

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

will contain **times** more grams of glucose than what you have in

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.