Answer
a) Enthalpy of hydration of $CaCl_2$: -2293 kJ/mol
Enthalpy of hydration of $CaI_2$: -2163 kJ/mol
b) $Cl^-$ is more strongly attracted to water than $I^-$.
Work Step by Step
a) The enthalpy of hydration equals the enthalpy change of solution plus the lattice energy. Using the values given, the enthalpy of hydration of $CaCl_2$ is $-46 kJ/mol + (-2247 kJ/mol) = -2293 kJ/mol$. Similarly, the enthalpy of hydration of $CaI_2$ is $-104 kJ/mol + (-2059 kJ/mol) = -2293 kJ/mol$.
b) The only thing that differs between $CaCl_2$ and $CaI_2$ is the $Cl$ and $I$ atoms. Using the enthalpies of hydration calculated in part (a), it can be determined that $CaCl_2$ is more strongly attracted to water than $CaI_2$ since a lower enthalpy of hydration suggests a greater affinity for water. Because of this, it follows that $Cl^-$ must be more strongly attracted to water than $I^-$.
