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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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Knowledge Check

  • The lattice energy of solid NaCl is 180 kcal mol^(-1) and enthaly of solution is 1 kcal mol^(-1) . If the hydration energies of Na^(+) and Cl^(-) ions are in the ratio 3:2, what is the enthalpy of hydration of sodium ion?

    A
    `-107.4" kcal "mol^(-1)`
    B
    `107.4" kcal "mol^(-1)`
    C
    `71.6" kcal "mol^(-1)`
    D
    `-71.6" kcal "mol^(-1)`

  • The lattice energy of solid NaCl is 180 kcal/mol. The dissolution of the solid in water in the form of ions is endothermic to the extent of 1 kcal/mol. If the hydration energies of Na^(+) and Cl^(-) are in the ratio 6: 5 , what is the enthalpy of hydration of Nation ?

    A
    `-8.5` kcal `mol^(-1)`
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    `-97.64` kcal `"mol"^(-1)`
    C
    `+82.6` kcal `"mol"^(-1)`
    D
    `+100` kcal `"mol"^(-1)`

  • 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) ) is

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    `+788`
    B
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    C
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    D
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