# 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

Jun 15, 2016

$\text{100 mL}$

#### 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 $\text{100 mL}$ of $\text{0.10 M}$ solution

$\textcolor{p u r p \le}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{c = {n}_{\text{solute"/V_"solution" implies n_"solute" = c * V_"solution}}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

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 $\text{0.98 g}$ of sulfuric acid in $\text{100 mL}$ 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 $\text{9.8 g}$ per liter, which means that its density is ${\text{9.8 g L}}^{- 1}$.

The target solution has a density of ${\text{4.9 g L}}^{- 1}$. 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 $\text{100 mL}$ fo solution and add $\text{100 mL}$ of water, you get

${m}_{\text{target" = "100 mL" + "100 mL" = "200 mL}}$

This $\text{200 mL}$ solution will still contain $\text{0.98 g}$ 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 $\text{4.9 g}$ of sulfuric acid per liter, so its density is now equal ${\text{4.9 g L}}^{- 1}$.

Therefore, the answer is (C) $\text{100 mL}$ of water.