1.

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.

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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