Question

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?

Answer 1

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 `"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

`"moles = molarity × litres"`

∴ `"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`

`"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 `"68.0 g H"_2"O"_2`.

Including the volume change

Let the mass of `"H"_2"O"_2 = xcolor(white)(l) "g"`

`"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`

`"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`

`"Total volume" = "1.500 L + 0.000 6932"x color(white)(l)"L"`

`"molarity" = "moles"/"litres"`

`(1.33 color(red)(cancel(color(black)("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"))))`

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

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

`1.995 +"0.000 9220"x = "0.029 40"x`

`"0.028 48"x = 1.995`

`x = 70.0`

Including the volume change, you need `70.0color(white)(l) "g H"_2"O"_2`.