# How many grams of hydrogen peroxide H_2O_2 must be added to 1,500 ml of water to produce a concentration of 1.33 M?

Apr 17, 2016

If there is no volume change, 68.0 g; if there is a volume change, 70.0 g.

#### Explanation:

You don't say if we are allowed to ignore the volume change on adding the ${\text{H"_2"O}}_{2}$, so I will give both answers.

Ignoring the volume change

The formula for molarity is

color(blue)(|bar(ul(color(white)(a/a)"molarity" = "moles"/"litres"color(white)(a/a)|)))" "

We can rearrange the formula to give

$\text{moles = molarity × litres}$

${\text{moles" =("1.33 mol H"_2"O"_2)/(1 color(red)(cancel(color(black)("L")))) × 1.500 color(red)(cancel(color(black)("L"))) = "2.00 mol H"_2"O}}_{2}$

${\text{mass" = 2.00 color(red)(cancel(color(black)("mol H"_2"O"_2))) × ("34.01 g H"_2"O"_2)/(1 color(red)(cancel(color(black)("mol H"_2"O"_2)))) = "68.0 g H"_2"O}}_{2}$

Ignoring dilution, you need ${\text{68.0 g H"_2"O}}_{2}$.

Including the volume change

Let the mass of $\text{H"_2"O"_2 = xcolor(white)(l) "g}$

${\text{Moles of H"_2"O"_2 = x color(red)(cancel(color(black)("g H"_2"O"_2))) × ("1 mol H"_2"O"_2)/(34.01 color(red)(cancel(color(black)("g H"_2"O"_2)))) = "0.02940"xcolor(white)(l) "mol H"_2"O}}_{2}$

${\text{Volume of H"_2"O"_2 = x color(red)(cancel(color(black)("g H"_2"O"_2))) × ("1 mL H"_2"O"_2)/(1.4425 color(red)(cancel(color(black)("g H"_2"O"_2)))) = 0.6932x color(white)(l) "mL H"_2"O"_2 = "0.000 6932"x color(white)(l)"L H"_2"O}}_{2}$

$\text{Total volume" = "1.500 L + 0.000 6932"x color(white)(l)"L}$

$\text{molarity" = "moles"/"litres}$

$\left(1.33 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{mol H"_2"O"_2))))/(1 color(red)(cancel(color(black)("L")))) = ("0.029 40" x color(red)(cancel(color(black)("mol H"_2"O"_2))))/(( "1.500 + 0.000 6932"x) color(red)(cancel(color(black)("L}}}}\right)$

$1.33 = \left(\text{0.029 40" x)/( "1.500 + 0.000 6932} x\right)$

1.33("1.500 + 0.000 6932"x) = "0.029 40"x

$1.995 + \text{0.000 9220"x = "0.029 40} x$

$\text{0.028 48} x = 1.995$

$x = 70.0$

Including the volume change, you need $70.0 \textcolor{w h i t e}{l} {\text{g H"_2"O}}_{2}$.