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Chapter 11: Problem 87

Calculate the freezing point and the boiling point of each of the following solutions. (Assume complete dissociation.) a. 5.0 \(\mathrm{g} \mathrm{NaCl}\) in 25 \(\mathrm{g} \mathrm{H}_{2} \mathrm{O}\) b. 2.0 \(\mathrm{g} \mathrm{Al}\left(\mathrm{NO}_{3}\right)_{3}\) in 15 \(\mathrm{g} \mathrm{H}_{2} \mathrm{O}\)

Short Answer

Expert verified
The freezing and boiling points of the solutions are as follows: a. For the 5.0 g NaCl solution in 25 g H2O: Freezing point: \(-12.7\text{ °C}\) and boiling point: \(103.5\text{ °C}\) b. For the 2.0 g Al(NO₃)₃ solution in 15 g H2O: Freezing point: \(-4.66\text{ °C}\) and boiling point: \(101.3\text{ °C}\)

Step by step solution

01

Calculate the molality of the NaCl solution

To calculate the molality of the NaCl solution, we will first convert the given solute and solvent masses to moles and kilograms: Moles of NaCl = \(\frac{5\text{g}}{58.44\text{g/mol}}\) = 0.0856 mol (using the molar mass of NaCl) Kilograms of water = \(\frac{25\text{g}}{1000\text{g/kg}}\) = 0.025 kg Now, we can calculate the molality: Molality (m) = \(\frac{0.0856\text{ mol}}{0.025\text{ kg}}\) = 3.42 mol/kg
02

Calculate the freezing point depression and boiling point elevation for NaCl solution

To calculate the freezing point depression and boiling point elevation for the NaCl solution, we will use the formulas above. Since NaCl is a strong electrolyte and dissociates into two ions (Na+ and Cl-), the van't Hoff factor (i) is 2: Freezing point depression: \(ΔT_f = K_f × m × i = 1.86\text{ °C/mol·kg} × 3.42\text{ mol/kg} × 2 = 12.7\text{ °C}\) Boiling point elevation: \(ΔT_b = K_b × m × i = 0.512\text{ °C/mol·kg} × 3.42\text{ mol/kg} × 2 = 3.50\text{ °C}\)
03

Calculate the new freezing and boiling points of the NaCl solution

Now, we can calculate the new freezing and boiling points by subtracting the freezing point depression and adding the boiling point elevation to the normal freezing and boiling points of water (0°C and 100°C, respectively): New freezing point = \(0\text{ °C} - 12.7\text{ °C} = -12.7\text{ °C}\) New boiling point = \(100\text{ °C} + 3.50\text{ °C} = 103.5\text{ °C}\) So, the freezing and boiling points of the 5.0 g NaCl solution in 25 g H2O are -12.7°C and 103.5°C, respectively. ##Solution B##
04

Calculate the molality of the Al(NO3)3 solution

To calculate the molality of the Al(NO₃)₃ solution, we will first convert the given solute and solvent masses to moles and kilograms: Moles of Al(NO₃)₃ = \(\frac{2\text{g}}{213.0\text{g/mol}}\) = 0.00939 mol (using the molar mass of Al(NO₃)₃) Kilograms of water = \(\frac{15\text{g}}{1000\text{g/kg}}\) = 0.015 kg Now, we can calculate the molality: Molality (m) = \(\frac{0.00939\text{ mol}}{0.015\text{ kg}}\) = 0.626 mol/kg
05

Calculate the freezing point depression and boiling point elevation for Al(NO3)3 solution

To calculate the freezing point depression and boiling point elevation for the Al(NO₃)₃ solution, we will use the same formulas. Since Al(NO₃)₃ is a strong electrolyte and dissociates into four ions (Al³+ and 3 NO₃⁻), the van't Hoff factor (i) is 4: Freezing point depression: \(ΔT_f = K_f × m × i = 1.86\text{ °C/mol·kg} × 0.626\text{ mol/kg} × 4 = 4.66\text{ °C}\) Boiling point elevation: \(ΔT_b = K_b × m × i = 0.512\text{ °C/mol·kg} × 0.626\text{ mol/kg} × 4 = 1.28\text{ °C}\)
06

Calculate the new freezing and boiling points of the Al(NO3)3 solution

Now, we can calculate the new freezing and boiling points by subtracting the freezing point depression and adding the boiling point elevation to the normal freezing and boiling points of water (0°C and 100°C, respectively): New freezing point = \(0\text{ °C} - 4.66\text{ °C} = -4.66\text{ °C}\) New boiling point = \(100\text{ °C} + 1.28\text{ °C} = 101.3\text{ °C}\) So, the freezing and boiling points of the 2.0 g Al(NO₃)₃ solution in 15 g H2O are -4.66°C and 101.3°C, respectively.

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Most popular questions from this chapter

The freezing-point depression of a \(0.091-m\) solution of \(\mathrm{CsCl}\) is \(0.320^{\circ} \mathrm{C} .\) The freezing-point depression of a \(0.091-\mathrm{m}\) solution of \(\mathrm{CaCl}_{2}\) is $0.440^{\circ} \mathrm{C} .$ In which solution does ion association appear to be greater? Explain.

What volume of ethylene glycol $\left(\mathrm{C}_{2} \mathrm{H}_{6} \mathrm{O}_{2}\right),$ a nonelectrolyte,must be added to 15.0 \(\mathrm{L}\) water to produce an antifreeze solution with a freezing point of \(-25.0^{\circ} \mathrm{C} ?\) What is the boiling point of this solution? (The density of ethylene glycol is \(1.11 \mathrm{g} / \mathrm{cm}^{3},\) and the density of water is 1.00 \(\mathrm{g} / \mathrm{cm}^{3} . )\)

Calculate the solubility of \(\mathrm{O}_{2}\) in water at a partial pressure of \(\mathrm{O}_{2}\) of 120 torr at \(25^{\circ} \mathrm{C}\) . The Henry's law constant for \(\mathrm{O}_{2}\) is $1.3 \times 10^{-3} \mathrm{mol} / \mathrm{L} \cdot\( atm for Henry's law in the form \)C=k P\( where \)C$ is the gas concentration \((\mathrm{mol} / \mathrm{L})\)

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.

If 500 . g of water is added to 75 \(\mathrm{g}\) of a \(2.5-m\) NaCl solution, what is the mass percent of \(\mathrm{NaCl}\) in the diluted solution?
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