# Question 9d8f3

May 15, 2017

$2 \cdot {10}^{23}$

#### Explanation:

The idea here is that the solution's mass by volume percent concentration, $\text{m/v%}$, 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 $\text{100 mL}$ of solution.

In your case, a $\text{9% m/v}$ glucose solution will contain $\text{9 g}$ of glucose, the solute, for every $\text{100 mL}$ of solution.

$\text{500 mL" = color(red)(5) * "100 mL}$

will contain $\textcolor{red}{5}$ times more grams of glucose than what you have in $\text{100 mL}$ 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"))))#

$= \textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{2 \cdot {10}^{23} \textcolor{w h i t e}{.} \text{molecules glucose}}}}$

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