# How do you calculate the molarity (M) of 157.6 g of #H_2SO_4# in 1.170 L of solution?

##### 1 Answer

#### Explanation:

An interesting approach to try here is to calculate the number of *grams* of sulfuric acid present in *moles* present in

So, you know that you get

#1 color(red)(cancel(color(black)("L solution"))) * ("157.6 g H"_2"SO"_4)/(1.170color(red)(cancel(color(black)("L solution")))) = "134.7 g H"_2"SO"_4#

So, you know that this solution contains

#("134.7 g H"_2"SO"_4)/"1 L solution"#

Now, **molarity** is defined as the number of **moles** of solute present in **molar mass** of sulfuric acid to convert it to *moles*

#134.7 color(red)(cancel(color(black)("g"))) * ("1 mole H"_2"SO"_4)/(98.079color(red)(cancel(color(black)("g")))) = "1.373 moles H"_2"SO"_4#

Therefore, you can say that the molarity of the solution is equal to

#color(darkgreen)(ul(color(black)("molarity = 1.373 mol L"^(-1))))#

The answer is rounded to four **sig figs**.