The lattice energy of solid `NaCI` is `180 kcal mol^(-1)`. The dissolution of the solid in `H_(2)O` is endothermic to the extent of `1.0 kcal mol^(-1)`. If the hydration energies of `Na^(o+)` and `CI^(Theta)` ions are in the ratio of `6:5` what is the enthalpy of hydration of sodium ion?

Options

A. `-85.6 kcal mol^(-1)`
B. `-97.64kcal mol^(-1)`
C. `+82.6kcal mol^(-1)`
D. `+100 kcal mol^(-1)`

Answer

B

Explanation

`DeltaH_(Solution)=DeltaH_(Lat tice)+DeltaH_(Hydration)`
`:. 1=180+DeltaH_(Hydration)`
`:. DeltaH_(Hydration)=-179kcal`
`DeltaH_(Na^(+))(hydration)+DeltaH_(Cl^(-))(hydration)=-179kcal`
`DeltaH_(Na^(+))+(5)/(6)DeltaH_(Na^(+))=-179kcal`
`DeltaH_(Na^(+))(hydration)=(6xx(-179))/(11)`
`=-97.64kcal mol^(-1)`