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

1 Answer
Jun 10, 2017

Answer:

#"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.