Question

How many moles of `H_2SO_4` are present in half a liter of a 0.150 M `H_2SO_4` solution?

Answer 1

`0.075*mol`

Explanation:

`"Molarity"="Moles of solute"/"Volume of solution"`

And thus, `"Moles of solute"="Volume of solution"xx"molarity"`

`=0.5*Lxx0.150*mol*L^-1=0.075*mol` with respect to `H_2SO_4`.

Of course, we speak of a formal concentration. We know that in aqueous solution sulfuric acid speciates to give `HSO_4^-` and `SO_4^(2-)`, and equivalent quantities of `H_3O^+`.