Question

What mass of lithium phosphate must be added to 25.0 L of water (Kf = 1.86 °C/m) to produce a solution that freezes at -5.2 °C and what is the molarity of the lithium ions in this solution when the density=2.34g/mL

Answer 1

. Cliven: Mass of water = 25.0 L x 10 = 25.0 kg Depression in freezing point, ATP = Tf -TE = 0°C -(-5.2°C) = 5.2 °C Suppose mass of Liz Pou required = x g Note: Lizpou H₂O, 3 Lit + PO, 3- - van't Hoff factor of Lizpou, i = 4 Now, ATF = ikem ; m = molality Now 1 x g 4 x 1.86 °C X → 5:2 °C = m . 115.79 g/mol 25.0 kg → 2 = 2023 so, mass of lithium phosphate = 2023. g) ☆ Mass of solution = 15.0 kg + 2.023 kg = 27.023 kg

Volume q solution - e mass Volume of solution = 27.023 x 103 ng - *** density 2.34 g ImL = 11548 mL Now, molarity of solution = 2023 g 115.79 g/mol 1548 mt. 1000 mL/L = Lit ions 1.51 mol/L per unit of Li3PO4, Since Liz Pou gives 3 Molarity of Lit ions = 3 x 1.51 mol/l = 4.54 M