# What is the molarity of 98% "H"_2"SO"_4?

Apr 28, 2017

Then that means you have ${\text{0.98 L H"_2"SO}}_{4}$ in $\text{1 L}$ solution, for instance. Let's assume we are at $\text{298.15 K}$. Then, assuming the density of 98% $\text{v/v}$ ${\text{H"_2"SO}}_{4}$ is approximately the same as pure ${\text{H"_2"SO}}_{4}$, we have:

$\text{0.98 L H"_2"SO"_4 xx "1.84 g"/"mL" xx "1000 mL"/"L" ~~ "1803.2 g}$

and the mols of it would be:

${\text{1803.2 g H"_2"SO"_4 xx "1 mol"/"98.079 g" = "18.39 mols H"_2"SO}}_{4}$

Therefore, the concentration is:

$\left(\text{18.39 mols H"_2"SO"_4)/("1 L solution") ~~ color(blue)("18.4 M}\right)$