Sulfuric acid has a molar mass of 98 g/mol. In the laboratory there is 100 mL of a 0.10 M sulfuric acid solution. How much water needs to be added to this volume to prepare a solution containing 4.9 g/L of sulfuric acid?

(A) 10 mL
(B) 50 mL
(C) 100 mL
(D) 150 mL
(E) 200 mL

1 Answer
Jun 15, 2016

Answer:

#"100 mL"#

Explanation:

Your strategy here will be to

  • use the volume and molarity of the initial solution to determine how many moles of sulfuric acid it contains
  • use the molar mass of sulfuric acid to convert the number of moles to grams
  • calculate the density of the initial solution and compare it with that of the target solution

So, calculate how many moles of sulfuric acid you have in #"100 mL"# of #"0.10 M"# solution

#color(purple)(|bar(ul(color(white)(a/a)color(black)(c = n_"solute"/V_"solution" implies n_"solute" = c * V_"solution")color(white)(a/a)|)))#

You will have

#n_("H"_2"SO"_4) = "0.10 mol" color(red)(cancel(color(black)("L"^(-1)))) * overbrace(100 * 10^(-3)color(red)(cancel(color(black)("L"))))^(color(blue)("volume in liters"))#

#n_("H"_2"SO"_4) = "0.010 moles H"_2"SO"_4#

This many moles are equivalent to

#0.010 color(red)(cancel(color(black)("moles H"_2"SO"_4))) * "98 g"/(1color(red)(cancel(color(black)("mole H"_2"SO"_4)))) = "0.98 g"#

Since you get #"0.98 g"# of sulfuric acid in #"100 mL"# of solution, it follows that one liter of this solution will contain

#10^3color(red)(cancel(color(black)("mL solution"))) * ("0.98 g H"_2"SO"_4)/(100color(red)(cancel(color(black)("mL solution")))) = "9.8 g H"_2"SO"_4#

Now, the initial solution contains #"9.8 g"# per liter, which means that its density is #"9.8 g L"^(-1)#.

The target solution has a density of #"4.9 g L"^(-1)#. Since you must make the target solution by adding water to the initial solution, i.e. by diluting it, you know for a fact that the mass of sulfuric acid remains unchanged.

This means that you can halve the density of the initial solution by doubling its volume.

If you start with #"100 mL"# fo solution and add #"100 mL"# of water, you get

#m_"target" = "100 mL" + "100 mL" = "200 mL"#

This #"200 mL"# solution will still contain #"0.98 g"# of sulfuric acid, which means that one liter will now contain

#10^3color(red)(cancel(color(black)("mL solution"))) * ("0.98 g H"_2"SO"_4)/(200color(red)(cancel(color(black)("mL solution")))) = "4.9 g H"_2"SO"_4#

The target solution contains #"4.9 g"# of sulfuric acid per liter, so its density is now equal #"4.9 g L"^(-1)#.

Therefore, the answer is (C) #"100 mL"# of water.