A student has a stock solution of 30.0% w/v hydrogen peroxide, #H_2O_2#, solution. How should the student prepare 250 mL of a 0.25% w/v #H_2O_2# solution?

1 Answer
Dec 1, 2015

Answer:

Here's how you could do that.

Explanation:

Your strategy here will be to figure out exactly how many grams of hydrogen peroxide are needed to make the target solution, then figure out what volume of the stock solution would contain that many grams.

Weight by volume percent concentration is defined as mass of solute, which in your case is hydrogen peroxide, divided by volume of solution, and multiplied by #100#.

#color(blue)("% w/v" = "mass of solute"/"volume of solution" xx 100)#

So, a #"0.25% w/v"# hydrogen peroxide solution will contain #"0.25 g"# of solute for every #"100m0 mL"# of solution. This means that your #"250-mL"# target solution will contain

#250 color(red)(cancel(color(black)("mL solution"))) * ("0.25 g H"_2"O"_2)/(100color(red)(cancel(color(black)("mL solution")))) = "0.625 g H"_2"O"_2#

Your focus now should shift to the stock solution. More specifically, you need to figure out what volume of the stock solution will contain this many grams of hydrogen peroxide

#"% w/v" = m_"solute"/V_"sol" xx 100 implies V_"sol" = m_"solute"/"%w/v" xx 100#

In your case, you will get

#V_"sol" = (0.625 color(red)(cancel(color(black)("g"))))/(30.0 color(red)(cancel(color(black)("g")))/"mL") xx 100 = "2.0833 mL"#

Rounded to two sig figs, the volume will be

#V = color(green)("2.1 mL")#

So, in order to prepare your solution, you need to take #"2.1 mL"# of stock solution and add enough water to get the final volume of the solution to #"250 mL"#.