The mass of water added to 100 g of 60% phosphoric acid to obtain a 40% solution is equal to.
| May 13, 2021 | education
Data: mрH3PO4 is the mass of the taken phosphoric acid solution (mрH3PO4 = 100%); ωрH3PO4 – initial mass fraction of orthophosphoric acid (ωрH3PO4 = 60% = 0.6); ωр1H3PO4 is the required mass fraction (ωр1H3PO4 = 40% = 0.4).
To find out the mass of water for diluting the initial solution of phosphoric acid, we use the formula: ωр1H3PO4 = mH3PO4 / mрH3PO4 = ωрH3PO4 * mрH3PO4 / mр1H3PO4 = ωрH3PO4 * mрH3PO4 / (mрH3H2PO4).
Let’s calculate: ωр1H3PO4 = ωрH3PO4 * mрH3PO4 / (mрH3PO4 + ΔmH2O).
0.4 = 0.6 * 100 / (100 + ΔmH2O).
(100 + ΔmH2O) * 0.4 = 60.
40 + 0.4ΔmH2O = 60.
0.4ΔmH2O = 60 – 40.
0.4ΔmH2O = 20.
ΔmH2O = 20 / 0.4 = 50 g.
Answer: The mass of added water is 50 g.
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