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Chapter 11: Problem 48

a. Use the following data to calculate the enthalpy of hydration for calcium chloride and calcium iodide. $$\begin{array}{lll}{\mathrm{CaCl}_{2}(s)} & {-2247 \mathrm{k} / \mathrm{mol}} & {-46 \mathrm{kJ} / \mathrm{mol}} \\ {\mathrm{Cal}_{2}(s)} & {-2059 \mathrm{k} / \mathrm{mol}} & {-104 \mathrm{kJ} / \mathrm{mol}}\end{array}$$ b. Based on your answers to part a, which ion, \(\mathrm{Cl}^{-}\) or \(\mathrm{I}^{-},\) is more strongly attracted to water?

Short Answer

Expert verified
The enthalpy of hydration for CaCl2 is 2201 kJ/mol and for CaI2 is 1955 kJ/mol. Therefore, the Cl- ion is more strongly attracted to water compared to the I- ion.

Step by step solution

01

Calculate the enthalpy of hydration for CaCl2

We are given: Enthalpy of solution (CaCl2) = -46 kJ/mol Lattice enthalpy (CaCl2) = -2247 kJ/mol Using the formula, we can find the enthalpy of hydration for CaCl2: Enthalpy of hydration (CaCl2) = Enthalpy of solution (CaCl2) - Lattice enthalpy (CaCl2) Enthalpy of hydration (CaCl2) = -46 kJ/mol - (-2247 kJ/mol) Enthalpy of hydration (CaCl2) = 2201 kJ/mol
02

Calculate the enthalpy of hydration for CaI2

We are given: Enthalpy of solution (CaI2) = -104 kJ/mol Lattice enthalpy (CaI2) = -2059 kJ/mol Using the formula, we can find the enthalpy of hydration for CaI2: Enthalpy of hydration (CaI2) = Enthalpy of solution (CaI2) - Lattice enthalpy (CaI2) Enthalpy of hydration (CaI2) = -104 kJ/mol - (-2059 kJ/mol) Enthalpy of hydration (CaI2) = 1955 kJ/mol
03

Compare the enthalpies of hydration and determine the ion more strongly attracted to water

We now have the enthalpies of hydration for both CaCl2 and CaI2: Enthalpy of hydration (CaCl2) = 2201 kJ/mol Enthalpy of hydration (CaI2) = 1955 kJ/mol Higher the enthalpy of hydration, stronger is the attraction between the ions and water. Since the enthalpy of hydration for CaCl2 is greater than that of CaI2, the Cl- ion is more strongly attracted to water as compared to the I- ion.

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Most popular questions from this chapter

An unknown compound contains only carbon, hydrogen, and oxygen. Combustion analysis of the compound gives mass percents of 31.57\(\% \mathrm{C}\) and 5.30\(\%\) H. The molar mass is determined by measuring the freezing-point depression of an aqueous solution. A freezing point of $-5.20^{\circ} \mathrm{C}\( is recorded for a solution made by dissolving 10.56 \)\mathrm{g}$ of the compound in 25.0 \(\mathrm{g}\) water. Determine the empirical formula, molar mass, and molecular formula of the compound. Assume that the compound is a nonelectrolyte.

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