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

Jun 10, 2017

${\text{1.373 mol L}}^{- 1}$

#### Explanation:

An interesting approach to try here is to calculate the number of grams of sulfuric acid present in $\text{1 L}$ of solution first, then convert this to moles present in $\text{1 L}$ of solution.

So, you know that you get $\text{157.6 g}$ of sulfuric acid, the solute, in $\text{1.170 L}$ of solution. Since solutions are homogeneous mixtures, you can use the known composition of the solution as a conversion factor to calculate the mass of sulfuric acid present in $\text{1 L}$ of solution

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 $\text{1 L}$ of solution. Since you already know the mass of solute present in the desired volume, use the 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

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{{\text{molarity = 1.373 mol L}}^{- 1}}}}$

The answer is rounded to four sig figs.