InterviewSolution
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Enthalpies of solution are given as follows :CuSO4(s) + 10H2O → CUSO4(10H2O)ΔH1 = -54.5 kJ mol-1CuSO4(s) + 100H2O → CUSO4(100H2O)ΔH2 = -68.4 kJ mol-1A solution contains 1 mol of CuSO4 in 180 g water at 25 °C. If it is diluted by adding 1620 g water, calculate the enthalpy of dilution. |
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Answer» Given : Enthalpy of solution of CuSO4 in 10 mol H2O = ΔsolnH = ΔH1 = -54.5 kJ mol-1 Enthalpy of solution of CuSO4 in 100 mol H2O = ΔH2 = - 68.4 kJmol-1 Mass of water = 1620 g For dilution, ΔdilH = ? Now, 80 g H2O = \(\frac{180}{18}\) = 10 mol H2O And, 1620 g H2O = \(\frac{1620}{18}\) = 90 mol H2O Hence for heat of dilution, CUSO4(10H2O) + 90H2O(I) → CUSO4(100H2O) ΔdilH = ? ∴ ΔdilH = ΔH2 -ΔH1 = - 68.4 – (54.5) = -13.9 kJmol-1 ∴ Heat of dilution = ΔdilH = -13.9 kJ mol-1 |
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