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

Oct 31, 2016

Under $500 \cdot g$ of $\text{phosphoric acid}$.
Here, we consider only the mass of ${H}_{3} P {O}_{4}$ used to prepare the solution, not its speciation.
"Mass of phosphoric acid"=521xx10^-3*Lxx9.30*mol*L^-1xx97.99*g*mol^-1=??*g.
Here $1 \cdot m L = 1 \times {10}^{-} 3 L$, i.e. $1 \cdot L = {10}^{3} \cdot m L$.