Dissolution of ionic solid in water is possible when:

H

lattice

> H

hydration

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H

lattice

= H

hydration

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H

hydration

: H

lattice = 2:1

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H

hydration

> H

lattice

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Solution

The correct option is C $H$ hydration $> H$ lattice
Solution:- (D) $H_{hydration} > H_{lattice}$
In order to dissolve an ionic solid, water molecules must break up the interactions between all of the ions in the solid. The heat of hydration $(H_{hydration})$ offsets the lattice energy $(H_{lattice})$ of an ionic solid to allow for solution formation to occur typically when $H_{hydration} > H_{lattice}$ .

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Similar questions

A l C l_{3}

= 5137 K J m o l^{- 1}

△ H h y d r a t i o n A l^{3 +} = - 4665 K J / m o l^{- 1}

△ H h y d r a t i o n C l^{-} = - 381 K J / m o l^{- 1}

N a C l (s)

Δ_{l a t t i c e} H = + 788 k J m o l^{- 1}

Δ_{s o l} H = + 4 k J m o l^{- 1}

$N a_{2} S O_{4}$ is soluble in water while $B a S o_{4}$ is insoluble. Which of the reason is correct about the above statement

Hydration energy ${Δ H}_{h y d r a} . (k J m o l^{- 1})$	Lattice energy $Δ H_{u} (k J m o l^{- 1})$
$L i^{+}$ -499	LiCl-840
$N a^{+}$ -390	NaCl-776
$K^{+}$ -305	KCl-703
$C l^{-}$ -382

Hydration energy ${Δ H}_{h y d r a} . (k J m o l^{- 1})$	Lattice energy $Δ H_{u} (k J m o l^{- 1})$
$L i^{+}$ -499	LiCl-840
$N a^{+}$ -390	NaCl-776
$K^{+}$ -305	KCl-703
$C l^{-}$ -382

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