ThermochemistryMedium
Question
The dissolution of CaCl2·6H2O in a large volume of water is endothermic to the extent of 3.5 kcal/mol. For the reaction, CaCl2(s) + 6H2O(l) → CaCl2·6H2O(s); ΔH is −23.2 kcal.
The heat of solution of anhydrous CaCl2 in large quantity of water will be
Options
A.−26.7 kcal mol–1
B.−19.7 kcal mol–1
C.19.7 kcal mol–1
D.26.7 kcal mol–1
Solution
$- 23.2 = \Delta_{Sol}H_{CaCl_{2}(s)} - \Delta_{sol}H_{CaCl_{2}.6H_{2}O(s)}$
$\therefore\Delta_{sol}H_{CaCl_{2}(s)} = ( - 23.2) + (3.5) = - 19.7\text{ kcal/mol}$
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