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The enthalpy of solution of sodium chloride is `4 kJ mol^(-1)` and its enthalpy of hydration of ion is `-784 kJ mol^(-1)`. Then the lattice enthalpy of `NaCl` (in `kJ mol^(-1)`) isA. `+"780 kJ mol"^(-1)`B. `+"394 kJ mol"^(-1)`C. `+"788 kJ mol"^(-1)`D. `+"398 kJ mol"^(-1)` |
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Answer» Correct Answer - C `DeltaH_("sol")=DeltaH_("lattice")+DeltaH_("hyd")` `4=DeltaH_("lattice")+(-784)` `DeltaH_("lattice")=4-(-784)=+"788 kJ mol"^(-1)` |
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