Answer:
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Explanation:
6 Moles of H3PO4 weigh → 6x98=588 g/l of pure H3PO4 → if the initial concentration of H3PO4 is 85% (w/w) you have to weigh 588/0.85 = 691.7 g/l
You feed 69.17 g of phosphoric acid 85% in one volumetric flask with a capacity of 100 ml and then you feed very very slowly water until to get the mark of the flask) at 100 ml. Unfortunately adding water to concentrated acids (sulfuric, phosphoric, etc) is very difficult due the strong exothermicity developed during the dilution.
Is better to work in a more safety way feeding in a becker around 30 ml of water and feed slowly 69.17 g of phosphoric acid 85%. The volume of the obtained solution will be (69.17+30)/1.75 = 55.67 ml. Then you transfer the obtained solution into a 100 ml volumetric flask and you wash for 3 time the becker using each time 10 ml of water. In the flask you will have around 85–90 ml of solution. Feed the volumetric flask to the mark (100 ml) and the wished solution will be ready.