Question #59aee

2 Answers
Oct 23, 2015



Use M1V1=M2V2
M1=0.25M V1=25mL M2=0.12M

#(0.25*25)/0.12#= 52.08mL

Oct 23, 2015

#"27 mL"#


Here's how you can think about what's going on.

You have a #"25-mL"# sample of a #"0.25-M"# sulfuric acid solution, and want to know how much water must be added to this solution to get its concentration down to #"0.12 M"#.

As you know, molarity is defined as number of moles of solute, in your case sulfuric acid, divided by the volume of the solution - expressed in liters.

#C = n/V#

When you perform a dilution, you essentially keep the number of moles of solute constant, and increase the total volume of the solution.

In your case, this tells you that both solutions will have the same number of moles of sulfuric acid.

So, how many moles of sulfuric acid you get in your initial solution?

#n = C * V = "0.25 M" * 25 * 10^(-3)"L" = "0.00625 moles"#

What volume of the second solution would contain the same number of moles of sulfuric acid?

#V = n/C = (0.00625color(red)(cancel(color(black)("moles"))))/(0.12color(red)(cancel(color(black)("moles")))/"L") = "0.05208 L" = "52.08 mL"#

So, how much water must do you need to add to the first solution to get its volume to #"52.08 mL"#?

#V_"add" = "52.08 mL" - "25 mL" = "27.08 mL"#

Rounded to two sig figs, the answer will be

#V_"add" = color(green)("27 mL")#