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The lattice energy of solid KCl is 181 k...

The lattice energy of solid `KCl` is `181 kcal mol^(-1)` and the enthalpy of solution of `KCl` in `H_(2)O` is `1.0 kcal mol^(-1)`. If the hydration enthalpies of `K^(o+)` and `Cl^(Theta)` ions are in the ratio of `2:1` then the enthalpy of hydration of `K^(o+)` is `-20xK cal mol^(-1)`. Find the value of `x`.

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To solve the problem, we will use the given data about the lattice energy of KCl, the enthalpy of solution, and the ratio of hydration enthalpies of K⁺ and Cl⁻ ions. ### Step-by-Step Solution: 1. **Identify the Given Data**: - Lattice energy of KCl (ΔH_lattice) = 181 kcal/mol - Enthalpy of solution of KCl in water (ΔH_solution) = 1 kcal/mol - Hydration enthalpies of K⁺ and Cl⁻ ions are in the ratio of 2:1. ...
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