At $25^{\circ} C$ , $1$ mole of $M g S O_{4}$ was dissolved in water, the heat evolved was found to be $91.2 kJ$ . One mole of $M g S O_{4} .7 H_{2} O$ on dissolution gives a solution of the same composition accompanied by an absorption of $13.8 kJ$ . The enthalpy of hydration, i.e., $Δ H$ for the reaction
$M g S O_{4} (s) + 7 H_{2} O (l) \to M g S O_{4} .7 H_{2} O (s)$ is:

- 105 kJ/mol

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- 77.4 kJ/mol

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105 kJ/mol

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None of these

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Solution

The correct option is A $- 105 kJ/mol$
Given that
$M g S O_{4} (s) + n H_{2} O \to M g S O_{4} . n H_{2} O$ ;
$Δ_{r} H_{1} = - 91.2 kJ/mol \dots \dots (i)$
$M g S O_{4} .7 H_{2} O (s) + (n - 7) H_{2} O \to M g S O_{4} . n H_{2} O$
$Δ_{r} H_{2} = 13.8 kJ/mol \dots \dots (i i)$

$M g S O_{4} (s) + 7 H_{2} O (l) \to M g S O_{4} .7 H_{2} O (s)$
The heat of hydration for the above reaction is
$Δ H_{hyd} = Δ_{r} H_{1} - Δ_{r} H_{2}$
$= - 91.2 kJ/mol - 13.8 kJ/mol$
$= - 105 kJ/mol$

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