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Given that the hydration energies of Ca2+(g) and Cl-(g) are -1562KJ/Mole and -364KJ/Mole respectively. The heat of solution (Hsoln) for one Mole of CaCl2 is -82.9 KJ/Mole. Determine the lattice energy for CaCl2.

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∆Hydr of CaCl2 = ∆Hydr Ca2+ +∆Hydr Cl-
=-1562+(-364x2)
=-1562+-728
=-2290Kjmol-1
∆Hsol = ∆Hlaq+∆Hydr
∆Hlath = ∆Hsol-∆Hydr
= -82.9 = (-2290)
= -82.9+2290
= +2207.1Kj/mole-1 

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