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Chapter 11: Problem 119

The lattice energy of \(\mathrm{NaCl}\) is \(-786 \mathrm{kJ} / \mathrm{mol},\) and the enthalpy of hydration of 1 mole of gaseous Na' and 1 mole of gaseous \(\mathrm{Cl}^{-}\) ions is \(-783 \mathrm{kJ} / \mathrm{mol}\) . Calculate the enthalpy of solution per mole of solid NaCl.

Short Answer

Expert verified
The enthalpy of solution per mole of solid NaCl is -1569 kJ/mol, calculated using the formula: Enthalpy of solution = Lattice energy + Enthalpy of hydration, and plugging in the given values: -786 kJ/mol for lattice energy and -783 kJ/mol for enthalpy of hydration.

Step by step solution

01

Identify the given values

We have been given the lattice energy of NaCl (-786 kJ/mol) and the enthalpy of hydration of 1 mole of gaseous Na^+ and Cl^- ions (-783 kJ/mol).
02

Use the formula for enthalpy of solution

The formula for the enthalpy of solution is: Enthalpy of solution = Lattice energy + Enthalpy of hydration
03

Insert the given values into the formula

Now, we need to plug the given values into the formula: Enthalpy of solution = -786 kJ/mol + (-783 kJ/mol)
04

Calculate the enthalpy of solution

Perform the calculation: Enthalpy of solution = -1569 kJ/mol So, the enthalpy of solution per mole of solid NaCl is -1569 kJ/mol.

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