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Chapter 11: Problem 111

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

Short Answer

Expert verified
The enthalpy of solution for one mole of solid NaCl is \(-1569 \mathrm{~kJ/mol}\), calculated using the thermodynamic equation: \(ΔH_{solution} = ΔH_{lattice} + ΔH_{hydration}\), where \(ΔH_{lattice} = -786 \mathrm{~kJ/mol}\) and \(ΔH_{hydration} = -783 \mathrm{~kJ/mol}\).

Step by step solution

01

Write down the given values

We know the given values of lattice energy and enthalpy of hydration: ΔH_lattice = -786 kJ/mol (for NaCl) ΔH_hydration = -783 kJ/mol (for 1 mole of Na+ and 1 mole of Cl- ions)
02

Use the thermodynamic equation

Now we will use the following equation to find the enthalpy of solution: ΔH_solution = ΔH_lattice + ΔH_hydration
03

Substitute the given values in the equation and calculate ΔH_solution

Substitute the given values in the equation: ΔH_solution = (-786 kJ/mol) + (-783 kJ/mol) Now, add the values: ΔH_solution = -1569 kJ/mol The enthalpy of solution for one mole of solid NaCl is -1569 kJ/mol.

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Most popular questions from this chapter

Plants that thrive in salt water must have internal solutions (inside the plant cells) that are isotonic with (have the same osmotic pressure as) the surrounding solution. A leaf of a saltwater plant is able to thrive in an aqueous salt solution (at \(25^{\circ} \mathrm{C}\) ) that has a freezing point equal to \(-0.621^{\circ} \mathrm{C}\). You would like to use this information to calculate the osmotic pressure of the solution in the cell. a. In order to use the freezing-point depression to calculate osmotic pressure, what assumption must you make (in addition to ideal behavior of the solutions, which we will assume)? b. Under what conditions is the assumption (in part a) reasonable? c. Solve for the osmotic pressure (at \(25^{\circ} \mathrm{C}\) ) of the solution in the plant cell. d. The plant leaf is placed in an aqueous salt solution (at \(25^{\circ} \mathrm{C}\) ) that has a boiling point of \(102.0^{\circ} \mathrm{C}\). What will happen to the plant cells in the leaf?

In lab you need to prepare at least \(100 \mathrm{~mL}\) of each of the following solutions. Explain how you would proceed using the given information. a. \(2.0 \mathrm{~m} \mathrm{KCl}\) in water (density of \(\mathrm{H}_{2} \mathrm{O}=1.00 \mathrm{~g} / \mathrm{cm}^{3}\) ) b. \(15 \% \mathrm{NaOH}\) by mass in water \(\left(d=1.00 \mathrm{~g} / \mathrm{cm}^{3}\right)\) c. \(25 \% \mathrm{NaOH}\) by mass in \(\mathrm{CH}_{3} \mathrm{OH}\left(d=0.79 \mathrm{~g} / \mathrm{cm}^{3}\right)\) d. 0. 10 mole fraction of \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\) in water \(\left(d=1.00 \mathrm{~g} / \mathrm{cm}^{3}\right)\)

Which ion in each of the following pairs would you expect to be more strongly hydrated? Why? a. \(\mathrm{Na}^{+}\) or \(\mathrm{Mg}^{2+}\) d. \(\mathrm{F}^{-}\) or \(\mathrm{Br}^{-}\) b. \(\mathrm{Mg}^{2+}\) or \(\mathrm{Be}^{2+}\) e. \(\mathrm{Cl}^{-}\) or \(\mathrm{ClO}_{4}^{-}\) c. \(\mathrm{Fe}^{2+}\) or \(\mathrm{Fe}^{3+}\) f. \(\mathrm{ClO}_{4}^{-}\) or \(\mathrm{SO}_{4}^{2-}\)

How would you prepare \(1.0 \mathrm{~L}\) of an aqueous solution of sodium chloride having an osmotic pressure of 15 atm at \(22^{\circ} \mathrm{C}\) ? Assume sodium chloride exists as \(\mathrm{Na}^{+}\) and \(\mathrm{Cl}^{-}\) ions in solution.

Rubbing alcohol contains \(585 \mathrm{~g}\) isopropanol \(\left(\mathrm{C}_{3} \mathrm{H}_{7} \mathrm{OH}\right)\) per liter (aqueous solution). Calculate the molarity.
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