# Question #cd8a6

##### 1 Answer

#### Answer:

#### Explanation:

The mass of sulfuric acid present in your sample can be calculated by using the compound's **molar mass**.

#M_ ("M H"_ 2"SO"_ 4) = "98.079 g mol"^(-1)#

So, you know that **mole** of sulfuric acid has a mass of **moles** of sulfuric acid, so set up a conversion factor that has the mass of **mole** of sulfuric acid on top and

This will get you

#0.0015 color(red)(cancel(color(black)("moles H"_2"SO"_4))) * "98.079 g"/(1color(red)(cancel(color(black)("mole H"_2"SO"_4)))) = "0.15 g"#

To find the **molarity** of the solution, simply divide the number of moles it contains by the volume of the solution **in liters**. Since the mass of sulfuric acid is quite small, you can assume that the volume of the solution will be equal to the volume of water.

#["H"_2"SO"_4] = "0.0015 moles"/(100 * 10^(-3) quad "L") = "0.015 mol L"^(-1)#

You should round the answer to one significant figure because that's how many sig figs you have for the volume of the solution.