Question
At $${25^ \circ }C,$$ when $$1\,mole$$ of $$MgS{O_4}$$ was dissolved in water, the heat evolved was found to be $$91.2\,kJ.$$ One $$mole$$ of $$MgS{O_4}.$$ $$7{H_2}O$$ on dissolution gives a solution of the same composition accompanied by an absorption of $$13.8\,kJ.$$ The enthalpy of hydration, i.e., $$\Delta {H_h}$$ for the reaction
$$MgS{O_4}\left( s \right) + 7{H_2}O\left( l \right) \to MgS{O_4}.7{H_2}O\left( s \right)$$ is :
A.
$$ - 105\,kJ/mol$$
B.
$$ - 77.4\,kJ/mol$$
C.
$$105\,kJ/mol$$
D.
$${\text{None of these}}$$
Answer :
$$ - 105\,kJ/mol$$
Solution :
$$\eqalign{
& {\text{Given that}} \cr
& MgS{O_4}\left( s \right) + n{H_2}O \to MgS{O_4}\,\,n{H_2}O; \cr
& {\Delta _r}{H_1} = - 91.2\,kJ/mol\,\,\,...\left( {\text{i}} \right) \cr
& MgS{O_4}.7{H_2}O\left( s \right) + \left( {n - 7} \right){H_2}O \to MgS{O_4}\left( {n{H_2}O} \right) \cr
& {\Delta _r}{H_2} = 13.8\,kJ/mol\,\,\,\,...\left( {{\text{ii}}} \right) \cr
& {\text{or}}\,\,\Delta {{\text{H}}_{hyd}} = {\Delta _r}{H_1} - {\Delta _r}{H_2}\,{\text{equation }}\left( {\text{i}} \right){\text{ - }}\left( {{\text{ii}}} \right) \cr
& = - 91.2\,kJ/mol - 13.8\,kJ/mol \cr
& = - 105\,kJ/mol \cr} $$