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
$M g S O_{4} (s) + n H_{2} O \to M g S O_{4} (n H_{2} O)$ ;
$Δ_{r} H_{1} = - 91.2 kJ/mol \dots \dots (i)$
$M g S O_{4} .7 H_{2} O (s) + (n - 7) H_{2} O \to M g S O_{4} (n H_{2} O)$
$Δ_{r} H_{2} = 13.8 kJ/mol \dots \dots (i i)$
Equation (i) - (ii)
or $Δ H_{hyd} = Δ_{r} H_{1} - Δ_{r} H_{2}$
$= - 91.2 kJ/mol - 13.8 kJ/mol$
$= - 105 kJ/mol$

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Similar questions

Q. 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:

At 298 K, the heat of solution of MgSO $_{4}$ (s) is -91.21 KJ mol $^{- 1}$ and that of MgSO $_{4}$ . 7 H $_{2}$ O (s) is 13.81 KJ mol $^{- 1}$ . Calculate heat of hydration of MgSO $_{4}$ i.e., $Δ$ H for the reaction $M g S O_{4} (s) + 7 H_{2} O (l) \to M g S O_{4} .7 H_{2} O (s)$