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

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Answer 1 · 2016-10-31T08:34:34

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

Here, we consider only the mass of $H_3PO_4$ 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 $1*mL=1xx10^-3L$ , i.e. $1*L=10^3*mL$ .

How many grams of H_3PO_4H_3PO_4 are in 521 mL of a 9.30 M solution of H_3PO_4H_3PO_4?