Question

Calculate the mole fraction of phosphoric acid (H3PO4) in a 25.4% (by mass) aqueous solution. (Assume 750 mL of solution.) What is the molarity of the solution? What is the molality? (At 20 ° C, the density of phosphoric acid is 1.1462 g/mL and the density of water is 0.99823 g/mL.)

Answer 1

Answer let we have 100 g aquous solution. then then 100g 25.4g H₂Pou Moleculare = (3+1) +31 +(16x4) = 984 74.69 H₂O Molecular mass - 6x1) +16 = 18 u mass moles of Ho = 74-6 . moles of H₂PO = mass line gram) A molar mass = 25.4 98 = 0.259 mol = 4.144 mol Mole fraction H₂PO = moles of H₂PO4 moles of HPO & moles of water 0.259. 0. 2594 4.144 0.259 4.403 is 0.06

Molarity of He pou = moles of H₃PO4 x 1000 volume of solution (in ml) 0.259 x1000 750 = Molality of H₂PO = 259 0.345 M 750 moles of H₂PO4 x 100 mass of H₂O ling) 0.259 X1000 74.6 s 3.472 m .
unit of molarity = mol/litre

Unit of molality = mol/kg

In question mass percentage is given so I assumed we have total 100 gram solution and the calculations done accordingly.