Question

One mole of anhydrous $MgC{l}_2$ dissolves in water and librates $25{\text{cal.mol}}^{-1}$ of heat.
$\Delta H_{\text{hydration}}ofMgC{l}_2=-30{\text{cal.mol}}^{-1}$ The lattice energy of $MgC{l}_2$ is:

• A. $+5{\text{cal.mol}}^{-1}$
• B. $-5{\text{cal.mol}}^{-1}$
• C. $55{\text{cal.mol}}^{-1}$
• D. $-55{\text{cal.mol}}^{-1}$

Answer 1

The correct option is A $+5{\text{cal.mol}}^{-1}$

Given,
$\Delta H_{\text{dissolution}}=-25\text{kcal/mol}$
$\Delta H_{\text{hydration}}=30\text{kcal/mol}$

We know that,
$\Delta H_{\text{dissolution}}=\Delta H_{\text{lattice energy}}+\Delta H_{\text{hydration}}$
Hence, $\Delta H_{\text{lattice energy}}=-25+30=5{\text{kcal.mol}}^{-1}$