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Lattice energy of `NaCl_((s))` is `-788kJ mol^(-1)` and enthalpy of hydration is `-784kJ mol^(-1)`. Calculate the heat of solution of `NaCl_((s))`.A. `-4kJ`B. `4KJ`C. `-6KJ`D. `2KJ` |
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Answer» Correct Answer - B heat of solution = L.E+H.E |
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