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,
$M g S O_{4} (s) + n H_{2} O \to M g S O_{4} (n H_{2} O);$
$Δ_{1} H_{1} = - 91.2 kJ/mol . . . (i)$
$M g S O_{4} .7 H_{2} O (s) + (n - 7) H_{2} O$
$\to M g S O_{4} . n H_{2} O$
$Δ_{1} H_{2} = 13.8 kJ/mol . . . (i i)$
Equation $(i) - (i i)$
$\Rightarrow Δ H_{h y d} = Δ_{1} H_{1} - Δ_{1} H_{2}$
$= - 91.2 kJ/mol - 13.8 kJ/mol = - 105 kJ/mol$

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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 $M g S O_{4} (s) + 7 H_{2} O (l) \to M g S O_{4} .7 H_{2} O (s)$

Q. One mole of soda ash dissolves in water with evolution of 25 kJ of heat. Dissolution of 1 mol of the hydrate crystals in water resulted in the absorption of 67 kJ of heat. The heat released due to hydration of soda ash is

Q. One mole of soda ash dissolves in water with evoution of 25 kJ of heat. Dissolution of 1 mol of the hydragte crystals in water resulted in the absorption of 67 kJ of heat. The heat released due to hydration of soda ash is

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The correct option is A −105 kJ/molGiven that, MgSO4(s)+nH2O→MgSO4(nH2O); Δ1H1=−91.2 kJ/mol...(i) MgSO4.7H2O(s)+(n−7)H2O →MgSO4.nH2O Δ1H2=13.8 kJ/mol...(ii) Equation (i)−(ii) ⇒ΔHhyd=Δ1H1−Δ1H2 =−91.2 kJ/mol−13.8 kJ/mol=−105 kJ/mol