The lattice enthalpy of solid `NaCI` is `772 kJ mol^(-1)` and enthalpy of solution is `2kJ mol^(-1)`. If the hydration enthalpy of `Na^(+) &CI^(-)` ions are in the ratio of 3: 25, what is the magnitude of enthalpy of hydration (in kJ) of chloride ion ?

Answer

[350]

Explanation

[Lattice enthalpy `= 772 Kj//Mol`
`((DeltaH.E)_(Na^(+)))/((DeltaH.E)_(CI^(-))) = (30)/(25) = (6)/(5),underset((x))((DeltaHE)_(Na^(+))) =(6)/(5)underset((y))((DeltaHE)_(CI^(-)))`
`x = (6)/(5)y`
`(DeltaH)_(solution")=` Lattice enthalpy `+H.E.` of `na^(+) +H.E.` of `CI^(-)`
`2 = 272 +(y) +(6)/(5)(y)`
`- 770 = (11)/(5) y, y = (770 xx 5)/(11) = 350 KJ]`