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
(A) 10 mL
(B) 50 mL
(C) 100 mL
(D) 150 mL
(E) 200 mL
1 Answer
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
#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
#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
The target solution has a density of
This means that you can halve the density of the initial solution by doubling its volume.
If you start with
#m_"target" = "100 mL" + "100 mL" = "200 mL"#
This
#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
Therefore, the answer is (C)