# 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?

May 8, 2017

$\text{3.18 M}$

Knowing the molal concentration, here is what you know... by assuming $\text{1 kg}$ of aqueous solvent, we have:

• ${\text{3.58 mols H"_2"SO}}_{4}$

• $\text{1 kg water}$

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 $\text{mols solute"/"L solution}$.

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"

$= {1.35}_{112282} \times {10}^{3} \text{g soln}$

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")

$= {1.12}_{5935683}$ $\text{L soln}$

Therefore, its molar concentration is:

$\left(\text{3.58 mols H"_2"SO"_4)/(1.12_(5935683) "L soln}\right)$

$=$ $\textcolor{b l u e}{\text{3.18 M}}$