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?

1 Answer
Apr 17, 2016

Answer:

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#.