Question

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

Answer 1

Then that means you have `"0.98 L H"_2"SO"_4` in `"1 L"` solution, for instance. Let's assume we are at `"298.15 K"`. Then, assuming the density of `98%` `"v/v"` `"H"_2"SO"_4` is approximately the same as pure `"H"_2"SO"_4`, we have:

`"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:

`"1803.2 g H"_2"SO"_4 xx "1 mol"/"98.079 g" = "18.39 mols H"_2"SO"_4`

Therefore, the concentration is:

`("18.39 mols H"_2"SO"_4)/("1 L solution") ~~ color(blue)("18.4 M")`