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?

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.