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InterviewSolution
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One mole of anhydrous `MgCl_(2)` dissolves in water and librates `25 cal//mol` of heat. `Delta H_("hydration")` of `MgCl_(2)=30 cal//mol`. Heat of dissolution of `MgCl.H_(2)O`A. `+5` cal/molB. `-5` cal/molC. 55 cal/molD. `-55` cal/mol |
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Answer» Correct Answer - A `MgCl_(2).H_(2)O rarrMgCl_(2)(aq)Delta H_(1)= ?` `MgCl_(2)(aq)rarrMg^(2+)(aq)+2Cl^(-)(aq)Delta H_(2)= -30 cal//mol` `MgCl_(2)(s)rarrMg^(2+)(aq)+2Cl^(-1)(aq)Delta H_(3)= -25 cal//mol` `Delta H_(1)+Delta H_(2)=Delta H_(3)` `Delta H_(1)=Delta H_(3)-Delta H_(2)= -25+30= +5 cal//mol` |
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