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The dissolution of CaCl2.6H2O in a large volume of water is endothermic to the extent of 3.5Kcalmol1 . For the reaction, CaCl2(s)+6H2O(l)CaCl2.6H2O(s) ΔH is 23.2Kcal . The heat of solution of anhydrous CaCl2 in large quantity of water will be:
 A. ΔH=16.7Kcalmol1
 B. ΔH=19.7Kcalmol1
 C. ΔH=19.7Kcalmol1
 D. ΔH=16.7Kcalmol1

Answer
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Hint: Dissolution is defined as the process where a substance (solute) in any phases (solid, liquid or gaseous) dissolves in a suitable solvent to form a solution. Dissolution is differ from solubility (solubility is the maximum amount of solute that can dissolve in a solution at a given temperature).

Complete step by step solution:
In the question it given that the dissolution of CaCl2.6H2O in a large volume of water is endothermic to the extent of 3.5Kcalmol1 and for the reaction, CaCl2(s)+6H2O(l)CaCl2.6H2O(s) ΔH is 23.2Kcal .
Now, we have heat and change in enthalpy is 3.5Kcalmol1 and 23.2Kcal , let’s calculate the heat of solution which is asked in the question.
We know heat of solution is equal to the sum of change in enthalpy and heat. Therefore we have a required formula to calculate heat of solution.
Heat of solution = Change in enthalpy + Heat
Heat of solution (ΔH) =ΔH1+q
Where (ΔH) represents heat of solution, ΔH1 is change in enthalpy and q represents the heat.
By putting the values of known quantities in the above formula to calculate the heat of solution.
Heat of solution =23.2+3.5
On solving, we get heat of solution
Heat of solution (ΔH) =19.7Kcal

Note:
The dissolution process can be considered to occur in three steps. In the first step separate particles of the solute from each other, in the second step separate particles of the solute from each other and in the last step combine separated solute and solvent particles to make a solution.Enthalpy is a property of the thermodynamic system; enthalpy is the sum of the internal energy and the product of the pressure and volume of a thermodynamic system.