Question

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

Answer 1

`"1.373 mol L"^(-1)`

Explanation:

An interesting approach to try here is to calculate the number of grams of sulfuric acid present in `"1 L"` of solution first, then convert this to moles present in `"1 L"` of solution.

So, you know that you get `"157.6 g"` of sulfuric acid, the solute, in `"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 `"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 `"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

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

The answer is rounded to four sig figs.