# 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?

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(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

#### Answer:

#### Explanation:

Your strategy here will be to

use thevolumeandmolarityof the initial solution to determine how manymolesof sulfuric acid it containsuse themolar massof sulfuric acid to convert the number of moles togramscalculate thedensityof 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 **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 **per liter**, which means that its density is

The target solution has a density of **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

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

This **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 **per liter**, so its density is now equal

Therefore, the answer is **(C)**