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Dissolution of an ionic solid XY in water is endothermic to an extent of 4 kJ per mol. If lattice enthalpy of XY(s) is + 788 kJ per mol, then hydration enthalpy of $Y^{-}$ would be (the ratio of hydration enthalpy of X : Y is 5 : 4)

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– 435.55 kJ

– 394 kJ

– 549.9 kJ

– 348.45 kJ

answer is C.

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Detailed Solution

${ΔH}_{Solution} = {ΔH}_{lattice} + {ΔH}_{Hyd}$

$+ 4 kJ = + 788 + {ΔH}_{Hyd}$

${ΔH}_{Hyd} = - 784 kJ$

Hydration enthalpy of

$Y^{-} = \frac{4}{9} \times (784) = - 348.45 kJ$

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