Question

How many grams of `H_3PO_4` are in 521 mL of a 9.30 M solution of `H_3PO_4`?

Answer 1

Under `500*g` of `"phosphoric acid"`.

Here, we consider only the mass of `H_3PO_4` used to prepare the solution, not its speciation.

Explanation:

`"Mass of phosphoric acid"=521xx10^-3*Lxx9.30*mol*L^-1xx97.99*g*mol^-1=??*g.`

Here `1*mL=1xx10^-3L`, i.e. `1*L=10^3*mL`.