# A solution of #H_2SO_4(aq)# with a molal concentration of 3.58 m has a density of 1.200 g/mL. What is the molar concentration of this solution?

##### 1 Answer

#"3.18 M"#

Knowing the molal concentration, here is what you know... by assuming

#"3.58 mols H"_2"SO"_4#

#"1 kg water"#

By using the density of the solution, along with the mass of the solution, we can get the volume of the solution. Then, *by definition*, the molarity is

#m_(soln) = 3.58 cancel("mols H"_2"SO"_4) xx ("98.079 g H"_2"SO"_4)/cancel("1 mol H"_2"SO"_4) + cancel"1 kg water" xx "1000 g"/cancel"1 kg"#

#= 1.35_(112282) xx 10^(3) "g soln"#

Therefore, its volume is:

#1.35_(112282) xx 10^(3) cancel"g soln" xx cancel"1 mL soln"/(1.200 cancel"g soln") xx "1 L soln"/(1000 cancel"mL soln")#

#= 1.12_(5935683)# #"L soln"#

Therefore, its **molar concentration** is:

#("3.58 mols H"_2"SO"_4)/(1.12_(5935683) "L soln")#

#=# #color(blue)("3.18 M")#