At $25^{\circ} \mathrm{C}$, when $1 \mathrm{~mole}$ of $\mathrm{MgSO}_{4}$ was dissolved in water, the heat…

At $25^{\circ} \mathrm{C}$, when $1 \mathrm{~mole}$ of $\mathrm{MgSO}_{4}$ was dissolved in water, the heat evolved was found to be $91.2 \mathrm{~kJ} .$ One mole of $\mathrm{MgSO}_{4} .7 \mathrm{H}_{2} \mathrm{O}$ on dissolution gives a solution of the same composition accompanied by an absorption of $13.8 \mathrm{~kJ}$. The enthalpy of hydration, i.e., $\Delta \mathrm{H}_{\mathrm{h}}$ for the reaction $\mathrm{MgSO}_{4}(\mathrm{~s})+7 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \longrightarrow \mathrm{MgSO}_{4} .7 \mathrm{H}_{2} \mathrm{O}(\mathrm{s})$ is:
  1. $-105 \mathrm{~kJ} / \mathrm{mol}$
  2. $-77.4 \mathrm{~kJ} / \mathrm{mol}$
  3. $105 \mathrm{~kJ} / \mathrm{mol}$
  4. None of these

Solution

Given that
$\mathrm{MgSO}_{4}(\mathrm{~s})+\mathrm{nH}_{2} \mathrm{O} ightarrow \mathrm{MgSO}_{4} \mathrm{nH}_{2} \mathrm{O} ;$
$\Delta_{\mathrm{r}} \mathrm{H}_{1}=-91.2 \mathrm{~kJ} / \mathrm{mol}$
$\mathrm{MgSO}_{4} .7 \mathrm{H}_{2} \mathrm{O}(\mathrm{s})+(\mathrm{n}-7) \mathrm{H}_{2} \mathrm{O}$
$\quad ightarrow \mathrm{MgSO}_{4}\left(\mathrm{nH}_{2} \mathrm{O}ight)$
$\Delta_{\mathrm{r}} \mathrm{H}_{2}=13.8 \mathrm{~kJ} / \mathrm{mol}$
or $\Delta \mathrm{H}_{\text {hyd }}=\Delta_{\mathrm{r}} \mathrm{H}_{1}-\Delta_{\mathrm{r}} \mathrm{H}_{2}$ equation (i) $-$ (ii)
$=-91.2 \mathrm{~kJ} / \mathrm{mol}-13.8 \mathrm{~kJ} / \mathrm{mol}$
$=-105 \mathrm{~kJ} / \mathrm{mol}$

Asked in: JEE-TOPICTESTS-CHEMISTRY

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