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Calculate heat of solution of NaCl from ...

Calculate heat of solution of NaCl from the following data
Hydration energy of `Na^(+) = -389kJ mol^(-1)`
Hydration energy of `Cl^(-) = -382kJ mol^(-1)`
Lattice energy of `NaCl = +776 kJ mol^(-)`

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The correct Answer is:
5
Hydration energy of BNaCI=Hydratin energy of `Na^(+)` hydration energy of `CI^(-)`
=-389 -382 =-771 kj/mol
=[-771-(-776)]=5kj/mole
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