# Question cd8a6

Jan 15, 2018

$\text{0.15 g}$

#### 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 $1$ mole of sulfuric acid has a mass of $\text{98.079 g}$ and that your solution contains $0.0015$moles of sulfuric acid, so set up a conversion factor that has the mass of $1$ mole of sulfuric acid on top and $1$ mole of the acid on the bottom.

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.