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The lattice enthalpy of solid NaCl is 77...

The lattice enthalpy of solid `NaCl` is `772 kJ mol^(-1)` and enthalpy of solution is `2 kJmol^(-1)`. If the hydration enthaply of `Na^(+)` & `Cl^(-)` ions are in the ratio of `3 : 2.5`, what is the enthalpy of hydration of chloride ion ?

A

`-140 kJmol^(-1)`

B

`-350 kJmol^(-1)`

C

`-351.81 mol^(-1)`

D

None

Text Solution

Verified by Experts

The correct Answer is:
B
`DeltaH_("solution")=DeltaH_("Hyd")(x)+DeltaH_("Lattice")`
`2=x+772`
`x=-770`
Enthalpy of hydration of `Cl^(-) =2.5/5.5 xx770=-350" kJ mol"^(-1)`
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