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Chapter 13: Problem 76

What is the freezing point of an aqueous solution that boils at \(105.0^{\circ} \mathrm{C} ?\)

Short Answer

Expert verified
The freezing point of the aqueous solution is -18.2 °C.

Step by step solution

01

Identify the formula for boiling point elevation and the given information

The formula for boiling point elevation is given by: ΔT_b = Kb * m Where: ΔT_b = boiling point elevation (increase in boiling point from the normal boiling point of the solvent) Kb = molal boiling point elevation constant (for water, Kb = 0.51 °C/mol) m = molality of the solution Here, we are given the boiling point of the aqueous solution (105.0 °C). We know the normal boiling point of water is 100.0 °C. So, we can find the boiling-point elevation.
02

Calculate boiling point elevation

The boiling point elevation is the difference between the boiling point of the solution and the normal boiling point of water: ΔT_b = 105.0 °C - 100.0 °C = 5.0 °C Now that we have the boiling point elevation, we can find the molality of the solution.
03

Calculate the molality of the solution

Using the boiling point elevation formula, we can rearrange it to solve for molality: m = ΔT_b / Kb Substitute the values into the formula: m = 5.0 °C / 0.51 °C/mol = 9.80 mol/kg Now, we have the molality of the solution.
04

Identify the freezing point depression formula and given information

The formula for freezing point depression is given by: ΔT_f = Kf * m Where: ΔT_f = freezing point depression (decrease in freezing point from the normal freezing point of the solvent) Kf = molal freezing point depression constant (for water, Kf = 1.86 °C/mol) m = molality of the solution (which we already found in Step 3) The normal freezing point of water is 0.0 °C. We can use the molality and the freezing point depression formula to find the freezing point depression.
05

Calculate freezing point depression

To find the freezing point depression, substitute the values into the formula: ΔT_f = Kf * m = (1.86 °C/mol) * (9.80 mol/kg) = 18.2 °C Now we have the freezing point depression.
06

Calculate the freezing point of the solution

Finally, to find the freezing point of the solution, subtract the freezing point depression from the normal freezing point of water: Freezing point of the solution = Normal freezing point - ΔT_f = 0.0 °C - 18.2 °C = -18.2 °C So, the freezing point of the aqueous solution is -18.2 °C.

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Most popular questions from this chapter

Carbon disulfide \(\left(\mathrm{CS}_{2}\right)\) boils at \(46.30^{\circ} \mathrm{C}\) and has a density of 1.261 \(\mathrm{g} / \mathrm{mL}\) . (a) When 0.250 \(\mathrm{mol}\) of a nondissociating solute is dissolved in 400.0 \(\mathrm{mL}\) of \(\mathrm{CS}_{2},\) the solution boils at \(47.46^{\circ} \mathrm{C} .\) What is the molal boiling-point-elevation constant for \(\mathrm{CS}_{2}\) ? (b) When 5.39 \(\mathrm{g}\) of a nondissociating unknown is dissolved in 50.0 \(\mathrm{mL}\) of \(\mathrm{CS}_{2},\) the solution boils at \(47.08^{\circ} \mathrm{C} .\) What is the molar mass of the unknown?

The normal boiling point of ethanol, \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH},\) is \(78.4^{\circ} \mathrm{C} .\) When 9.15 \(\mathrm{g}\) of a soluble nonelectrolyte is dissolved in 100.0 \(\mathrm{g}\) of ethanol at that temperature, the vapor pressure of the solution is \(7.40 \times 10^{2}\) torr. What is the molar mass of the solute?

Acetonitrile \(\left(\mathrm{CH}_{3} \mathrm{CN}\right)\) is a polar organic solvent that dissolves a wide range of solutes, including many salts. The density of a 1.80 \(\mathrm{M}\) LiBr solution in acetonitrile is 0.826 \(\mathrm{g} / \mathrm{cm}^{3} .\) Calculate the concentration of the solution in (a) molality, (b) mole fraction of LiBr, (c) mass percentage of \(\mathrm{CH}_{3} \mathrm{CN} .\)

The solubility of \(\mathrm{Cr}\left(\mathrm{NO}_{3}\right)_{3} \cdot 9 \mathrm{H}_{2} \mathrm{O}\) in water is 208 \(\mathrm{g}\) per 100 \(\mathrm{g}\) of water at \(15^{\circ} \mathrm{C}\) . A solution of \(\mathrm{Cr}\left(\mathrm{NO}_{3}\right)_{3} \cdot 9 \mathrm{H}_{2} \mathrm{O}\) in water at \(35^{\circ} \mathrm{C}\) is formed by dissolving 324 \(\mathrm{g}\) in 100 \(\mathrm{g}\) of water. When this solution is slowly cooled to \(15^{\circ} \mathrm{C},\) no precipitate forms. (a) Is the solution that has cooled down to \(15^{\circ}\) Cunsaturated, saturated, or supersaturated? (b) You take a metal spatula and scratch the side of the glass vessel that contains this cooled solution, and crystals start to appear. What has just happened? (c) At equilibrium, what mass of crystals do you expect to form?

Commercial concentrated aqueous ammonia is 28\(\% \mathrm{NH}_{3}\) by mass and has a density of 0.90 \(\mathrm{g} / \mathrm{mL} .\) What is the molarity of this solution?
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