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Chapter 10: Problem 79

Consider the following solutions: \(0.010 m \mathrm{Na}_{3} \mathrm{PO}_{4}\) in water \(0.020 \mathrm{m} \mathrm{CaBr}_{2}\) in water \(0.020 \mathrm{m} \mathrm{KCl}\) in water \(0.020 \mathrm{m}\) HF in water (HF is a weak acid.) a. Assuming complete dissociation of the soluble salts, which solution(s) would have the same boiling point as \(0.040 \mathrm{m}\) \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\) in water? \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\) is a nonelectrolyte. b. Which solution would have the highest vapor pressure at \(28^{\circ} \mathrm{C} ?\) c. Which solution would have the largest freezing-point depression?

Short Answer

Expert verified
a) The Na3PO4 and KCl solutions will have the same boiling point as the 0.040 m C6H12O6 solution. b) The HF solution will have the highest vapor pressure at 28掳C. c) The CaBr2 solution will have the largest freezing-point depression.

Step by step solution

01

Calculate the molality for each solution

Since the given molality values are provided directly, we don't need to make any calculations for molality. We already have the molality for each of the solutions: - 0.010 m Na3PO4 - 0.020 m CaBr2 - 0.020 m KCl - 0.020 m HF - 0.040 m C6H12O6 (reference)
02

Determine the Van't Hoff factor (i) for each solution

For non-electrolytes like C6H12O6, i=1. For electrolytes, i is equal to the number of ions produced on dissociation: - Na3PO4: 3 Na+ + PO4^3- 鈫� i = 4 - CaBr2: Ca^2+ + 2 Br- 鈫� i = 3 - KCl: K+ + Cl- 鈫� i = 2 - HF: weak acid, doesn't completely dissociate 鈫� i 鈮� 1
03

Calculate i脳molality for each solution

Multiply molality and the number of ions produced(i) for each solution: - Na3PO4: 0.010 m 脳 4 = 0.040 - CaBr2: 0.020 m 脳 3 = 0.060 - KCl: 0.020 m 脳 2 = 0.040 - HF: 0.020 m 脳 1 鈮� 0.020 - C6H12O6: 0.040 m 脳 1 = 0.040 (reference) a) The solutions which have the same i脳molality as the C6H12O6 solution will have the same boiling point. Both Na3PO4 and KCl solutions show the same i脳molality value (0.040) as C6H12O6. b) The highest vapor pressure will be produced by the solution with the lowest molality. In this case, the HF solution has the lowest i脳molality, so it will have the highest vapor pressure at 28掳C. c) The largest freezing-point depression will occur in the solution with the highest i脳molality. So, CaBr2 with i脳molality value of 0.060 will have the largest freezing-point depression. Answers: a) The Na3PO4 and KCl solutions will have the same boiling point as the 0.040 m C6H12O6 solution. b) The HF solution will have the highest vapor pressure at 28掳C. c) The CaBr2 solution will have the largest freezing-point depression.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Boiling Point Elevation
Boiling point elevation is a key property of solutions that results from the introduction of a solute to a solvent, causing the boiling point of the resulting solution to be higher than that of the pure solvent. This occurs because the solute particles disrupt the solvent's ability to vaporize, which requires additional energy, thereby increasing the boiling point.
The extent of boiling point elevation is determined by the equation:\[\Delta T_b = i \cdot K_b \cdot m\]where \(\Delta T_b\) is the boiling point elevation, \(K_b\) is the ebullioscopic constant of the solvent, \(m\) is the molality of the solution, and \(i\) is the van 't Hoff factor, which represents the number of particles the solute breaks into upon dissolving.
In the exercise, two solutions have an i\(\times\)molality equal to that of the reference C\(_{6}\)H\(_{12}\)O\(_{6}\) solution: \(Na_3PO_4\) and \(KCl\). Thus, they will exhibit the same boiling point elevation as the reference solution.
Vapor Pressure
Vapor pressure is the pressure exerted by a vapor in equilibrium with its liquid or solid phase at a given temperature. It is influenced by the presence of a solite; when a non-volatile solute is added, the vapor pressure of the solvent decreases. This happens because solute particles occupy space at the surface, reducing the number of solvent molecules available to escape into the vapor phase.
The fundamental principle behind this is Raoult's Law, given by:\[P_{solution} = X_{solvent} \cdot P_{solvent}^0\]where \(P_{solution}\) is the vapor pressure of the solution, \(X_{solvent}\) is the mole fraction of the solvent, and \(P_{solvent}^0\) is the vapor pressure of the pure solvent.
In the provided exercise scenario, the HF solution exhibits the lowest contribution to vapor pressure reduction due to its low i\(\times\)molality value, meaning it will have the highest vapor pressure at the given temperature of 28掳C.
Freezing-Point Depression
Freezing-point depression is another colligative property that describes the reduction of a solution's freezing point compared to that of the pure solvent. This occurs because the solute disrupts the structure necessary for the solvent to solidify, requiring a greater removal of thermal energy (cooling).The formula to calculate freezing-point depression is:\[\Delta T_f = i \cdot K_f \cdot m\]where \(\Delta T_f\) is the freezing point depression, \(K_f\) is the cryoscopic constant of the solvent, \(m\) is the molality of the solute, and \(i\) is the van 't Hoff factor.
For this exercise, the solution of \(CaBr_2\), with the highest i\(\times\)molality value (0.060), will show the largest freezing-point depression. This indicates that the \(CaBr_2\) solution will require a greater decrease in temperature to solidify, compared to other listed solutions.

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Most popular questions from this chapter

Reserpine is a natural product isolated from the roots of the shrub Rauwolfia serpentina. It was first synthesized in 1956 by Nobel Prize winner R. B. Woodward. It is used as a tranquilizer and sedative. When 1.00 g reserpine is dissolved in \(25.0 \mathrm{g}\) camphor, the freezing-point depression is \(2.63^{\circ} \mathrm{C}(K_{\mathrm{f}}\) for camphor is \(40 .^{\circ} \mathrm{C} \cdot \mathrm{kg} / \mathrm{mol})\) Calculate the molality of the solution and the molar mass of reserpine.

Which of the following will have the lowest total vapor pressure at \(25^{\circ} \mathrm{C} ?\) a. pure water (vapor pressure \(=23.8\) torr at \(25^{\circ} \mathrm{C}\) ) b. a solution of glucose in water with \(\chi_{\mathrm{C}_{\mathrm{s}} \mathrm{H}_{\mathrm{l} 2} \mathrm{O}_{\mathrm{s}}}=0.01\) c. a solution of sodium chloride in water with \(\chi_{\mathrm{NaCl}}=0.01\) d. a solution of methanol in water with \(\chi_{\mathrm{CH_{3}}, \mathrm{OH}}=0.2\) (Consider the vapor pressure of both methanol \([143\) torr at \(\left.25^{\circ} \mathrm{C}\right]\) and water.

Liquid A has vapor pressure \(x,\) and liquid B has vapor pressure y. What is the mole fraction of the liquid mixture if the vapor above the solution is \(30 . \%\) A by moles? \(50 . \%\) A? \(80 . \%\) A? (Calculate in terms of \(x\) and \(y .\) ) Liquid A has vapor pressure \(x,\) liquid B has vapor pressure y. What is the mole fraction of the vapor above the solution if the liquid mixture is \(30 . \%\) A by moles? \(50 . \%\) A? \(80 . \%\) A? (Calculate in terms of \(x\) and \(y .\) )

An aqueous solution containing 0.250 mole of Q, a strong electrolyte, in \(5.00 \times 10^{2} \mathrm{g}\) water freezes at \(-2.79^{\circ} \mathrm{C}\). What is the van't Hoff factor for Q? The molal freezing-point depression constant for water is \(1.86^{\circ} \mathrm{C} \cdot \mathrm{kg} / \mathrm{mol} .\) What is the formula of \(\mathrm{Q}\) if it is \(38.68 \%\) chlorine by mass and there are twice as many anions as cations in one formula unit of Q?

How would you prepare 1.0 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.
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