# Question bc6af

Mar 24, 2017

${\text{10. mL H"_2"O}}_{2}$

#### Explanation:

A solution's volume by volume percent concentration, $\text{v/v %}$, is a measure of the number of milliliters of solute present for every $\text{100 mL}$ of solution.

In other words, the volume by volume percent concentration tells you how many milliliters of solute you would have if you were to pick a $\text{100-mL}$ sample of a given solution.

In your case, you know that the solution has a 3.0% volume by volume concentration, which means that every $\text{100 mL}$ of this solution contain $\text{3.0 mL}$ of hydrogen peroxide, the solute.

Now, solutions are homogeneous mixtures, which means that the same the same composition throughout. This allows you to use the percent concentration as a conversion factor to determine how many milliliters of hydrogen peroxide you'd get in $\text{348 mL}$ of solution.

348 color(red)(cancel(color(black)("mL solution"))) * overbrace(("3.0 mL H"_2"O"_2)/(100color(red)(cancel(color(black)("mL solution")))))^(color(blue)("= 3.0% v/v")) = "10.44 g H"_2"O"_2

The answer must be rounded to two sig figs, the number of sig figs you have for the percent concentration.

Therefore, you will have

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{\text{volume H"_ 2"O"_ 2 = "10. mL}}}}$

Mar 24, 2017

10.44mL ${H}_{2} {O}_{2}$

#### Explanation:

The antiseptic ${H}_{2} {O}_{2}$ solution is 3.0% (v/v). This means that for every 100mL of the antiseptic solution, you have 3mL of the solute, ${H}_{2} {O}_{2}$.

3% (v/v) is the same thing as writing as (3mL (H_2O_2))/(100mL(solution)) * 100%# so,

If for every 100mL of soln you have 3mL of ${H}_{2} {O}_{2}$, then for 348 mL you would have:

$\frac{3 m L \left({H}_{2} {O}_{2}\right)}{100 m L \left(s o l u t i o n\right)}$ = $\frac{x m L}{348 m L \left(s o l u t i o n\right)}$

$\cancel{348} m L \left(s o l u t i o n\right) \cdot \frac{3 m L \left({H}_{2} {O}_{2}\right)}{\cancel{100} m L \left(s o l u t i o n\right)}$ = $x m L \left({H}_{2} {O}_{2}\right)$

Answer: 10.44mL ${H}_{2} {O}_{2}$