There is a series of properties solutions display that vary with the concentration of solute particles, often regardless of chemical composition. In other words, these properties tend to change pretty much with the concentration. Such properties are referred to as collative properties and include changes in the freezing point, boiling point, vapor pressure, and osmotic pressure.
A good deal of the time, the changes in behavior are over a somewhat narrow range of concentrations, but are predictable over that range. Notably, there are solutions that vary the same way over the entire range of concentrations; these are known as ideal solutions. The forces of interaction between solvent and solute molecules of the ideal solution are the same as between the molecules of each component when separated. During the formation of an ideal solution from the separate components, there are no volume changes (additive volumes with no gain or loss) and no enthalpy changes. Pairs of chemically similar substances, such as methanol (CH3OH) and ethanol (C2H5OH), or benzene (C6H6) and toluene (C7H8), form ideal solutions. Dissimilar substances, such as ethanol and benzene, form nonideal solutions.
The vapor pressure of the solvent reduces as the concentration of the solute increases (inverse relationship). Generally, the discussions of vapor pressure are with reference to the concentration of nonvolatile solutes and, therefore, we are dealing with the tendency of only the solvent to leave the solution. Experimentation has determined that dilute solutions of equal molality, using the same solvent and different nonelectrolytes (no dissociation) as solutes, show the same amount of vapor pressure lowering (depression) in every case.
Experimentation has revealed an interesting point about vapor pressure changes. Suppose we were to make up two solutions, solution I and solution II, from the same solvent and using solutes that are nonvolatile and nonelectrolytes. Let us take two identical portions of the solvent and place in two different containers. In the case of solution I, mix a solution using one solute to achieve a particular molality. In the case of solution II, mix a solution to the same molality, but use two different solutes to do so. The lowering of the vapor pressure of the solute is the same in both solutions.
Raoult’s law states that the vapor pressure lowering is proportional to the mole fraction of the solute (or, the vapor pressure is proportional to the mole fraction of the solvent.). In equation form:
or
In the second form, the vapor pressure of the solution has been identified with the vapor pressure of the solvent over the solution, since the solute is assumed to be nonvolatile.
Further, Raoult’s law can be applied when two volatile components are mixed. In systems of liquids that mix in all proportions to form ideal solutions, Raoult’s law in the form of the second equation applies to the partial pressure of each volatile component separately.
Raoult’s law is explained by the hypothesis that solute molecules at the surface of the liquid interfere with the escape of solvent molecules into the vapor phase. Because of the vapor pressure lowering, the boiling point of the solution is raised and the freezing point is lowered, as compared with the boiling and freezing points of the pure solvent.
On the cooling of most dilute solutions, the solvent begins to crystallize before the solute crystallizes. The temperature at which the solvent’s crystals first begin to appear is the freezing point of that solution. The freezing point of a solution is always lower than the freezing point of the pure solvent. In dilute solutions, the freezing-point lowering, ΔTf, is directly proportional to the number of moles solute (number of molecules) present. The change in freezing point is determined by subtracting the true freezing point of the solution from the freezing point of the pure solvent (ΔTf(solvent) – ΔTf(solution)). The equation for the change in freezing point is
where m is the molality of the solution (Chapter 12). If this equation were to be valid up to a concentration of 1 m, the lowering of the freezing point of a 1 m solution of any nonelectrolyte dissolved in the solvent would be kf, which is called the molal freezing-point constant or, more simply, the freezing-point constant. The value of kf is a property characteristic for each solvent.
EXAMPLE 1 The kf for water is 1.86°C/m (1.86 K/m), which can be expressed as 1.86°C · kg H2O/mol solute (1.86 K · kg H2O/mol solute) after some algebraic rearrangement. This means that a solution of 1 mol cane sugar (342 g, over 
lb) dissolved in 1 kg of water should freeze at –1.86° C.
The freezing-point drop suggested in Example 1 is correct if the relationship holds true for this concentrated solution. As mentioned above (and below), the laws are most accurate for low (dilute) concentrations. Further, some of these calculations would be more accurate if mole fraction were to be used; however, at low concentrations, the error involved in calculations based on molality is very low.
Compared to the boiling point of the pure solvent, dilute solutions containing that solvent boil at a higher temperature. In dilute solutions, the elevation of the boiling point is directly proportional to the number of moles solute (or molecules). The equation for the change in boiling point is
The molal boiling-point constant of the solvent is kb. As with kf, the numerical value is a property of the solvent alone and is independent of the solute, assuming a solute that is nonvolatile and a nonelectrolyte. Also assumed in the tables presenting kb values is that the pressure is 1 atm (standard pressure).
EXAMPLE 2 The molal boiling-point constant for water is 0.512°C/m. If 1 mol cane sugar (342 g/mol) is dissolved in 1 kg water, the solution will boil at 100.512°C, assuming standard pressure. By this relationship, a half of a mole of sugar (171 g) would boil at 100.256°C/m and 2 mol sugar (684 g) should boil at 101.024°C/m. However, these are not necessarily solutions that can be mixed; 684 g of sugar is nearly a pound and a half of sugar, which you are supposed to dissolve in a liter of water!
Note: It would be a good idea to memorize constants relating to water. It is an extremely important compound on this planet, necessary for life, and a very commonly used solvent. Many professors will assume you know them during a test!
Imagine a container with a solution that is separated from its pure solvent by a membrane that allows the solvent to pass through, but not the solute. The membrane that allows specific things through, but not others, is a semipermeable membrane. Since substances diffuse (tend to move from higher concentration to lower concentration), the solvent will diffuse through the semipermeable membrane into the solution (osmosis). Let us carry the concept one step further.
Imagine the solvent being separated from the solution by a semipermeable membrane and there is a tube rising from the portion of the container containing the solution, as shown in Fig. 14-1. The solution will increase in volume due to the incoming solvent (and also become diluted by the solvent), forcing some of the solution up the open tube. At some point in time, the weight of the solution exerting a weight downward will become equal to the pressure forcing the solution upward (osmotic pressure) and equilibrium will have been reached, after which there will be no further rise in the height of the solution in the tube. The osmotic pressure (measurable pressure at this point in the process) can be measured in the usual units of pressure, such as Pa, atm, psi, or torr.
Fig. 14-1
The osmotic pressure, π, of a dilute solution of a nonelectrolyte is given by an equation formally equivalent to the ideal gas law:
in which M is molarity and T is in the Kelvin scale. If R = 0.0821 L · atm/mol · K, the value for π will be in atm and it may be expressed in other units of pressure, as can R by conversion from atm to the desired unit.
All the laws discussed so far are valid for dilute solutions of nonelectrolytes. If there is a solute that is an electrolyte, the ions contribute independently to the effective molal (or molar) concentration. The ions interact and, therefore, the effects are not as large as predicted by the mathematical equations.
EXAMPLE 3 A solution containing 0.100 mol of KCl per kilogram of water freezes at –0.345°C. The freezing-point lowering is calculated using the number of independent particles. If KCl were to ionize completely the molality of particles would be 0.200 m, and the calculated freezing point would be lowered from 0°Cby
providing us with the theoretical freezing point of –0.372°C. This difference in freezing point tells us that the ions do not act completely independently of each other. As a point, it is noted that a 0.100 m BaCl2 solution freezes at –0.470°C, but the predicted (Ba2+ and 2Cl–, 3 ions) should be (3 × 1.86 × 0.1 = 0.558°C) a freezing point of –0.558°C.
When discussing ΔTf and ΔTb, the theoretical number of ions can be indicated with i as follows:
and the value of i, although taken to be a whole number, it is actually not in many cases. The value of i for KCl in Example 3 is actually 1.85, not the 2 we get by assuming independent K+ and Cl– ions. In the case of BaCl2, the actual value of i is 2.53, not 3.
For any dilute solution, whether electrolyte or nonelectrolyte, the deviations from any one of the laws of the dilute solution are equal to the deviations from any of the others on a fractional or percentage basis. That is,
in which symbols labeled with ° indicate values predicted by the laws of the dilute solution.
Henry’s law states that at constant temperature, the concentration of a slightly soluble gas in a liquid is directly proportional to the partial pressure of the gas in the space above the liquid. The concentration can be expressed in either mass or the number of moles of gas dissolved in a specific volume of liquid.
When a mixture of two gases is in contact with a solvent, the amount of each gas that is dissolved is the same as if it were present alone at a pressure equal to its own partial pressure in the gas mixture.
Henry’s law applies only to dilute solutions at low pressures. If the gas involved dissolves very well in the solvent, the reason is most likely that the gas is reacting with the solvent. For example, CO2 appears to dissolve rather well in water, but actually reacts to produce H2CO3. Solutions that behave in this manner do not follow Henry’s law well at all.
A solute distributes itself between two immiscible (substances that do not mix) solvents so that the ratio of its concentrations in dilute solutions in the two solvents is constant, regardless of the actual concentration in either solvent. The solvents will form layers, since they do not mix. In this situation, both concentrations are assumed to be on the same volumetric basis (e.g., mol/L).
EXAMPLE 4 For the distribution of iodine between ether and water at room temperature, the value of the constant is about 200. This means that
The value of this ratio of concentrations is called the distribution ratio or distribution coefficient. It is equal to the ratio of the solubilities (per unit volume) in the two solvents if saturated solutions in these solvents are dilute enough for the law of distribution to apply.
14.1. The freezing point of pure camphor is 178.4°C and its molal freezing-point constant, kf, is 40.0°C/m. Find the freezing point of a solution containing 1.50 g of a compound that has the molar mass (M) 125 g/mol dissolved in 35.0 g of camphor.
Since the solution for this problem calls for the calculation of ΔT, which is dependent on molal concentration, we must calculate the moles of solute.
Then, we can calculate the freezing-point lowering by substitution.
Freezing point of solution = (freezing point of solvent) – ΔTf = 178.4°C – 13.7°C = 164.7°C
The solution for the freezing-point lowering could be set up by substituting directly into the T equation after the expansion of m as:
14.2. A solution containing 4.50 g of a nonelectrolyte (i = 1) dissolved in 125 g of water freezes at –0.372°C. Calculate the molar mass, M, of the solute.
First compute the molality from the freezing-point equation.
Then, from the definition of molality, compute the number of moles solute, n(solute), in the sample.
And since
we solve for M:
14.3. The molar mass of a nonvolatile solute is 58.0. Calculate the boiling point of a solution containing 24.0 g of the solute dissolved in 600 g water. The barometric pressure is such that pure water boils at 99.725°C. Note: i is expected to be 1, since there is no mention of disassociation.
Elevation of boiling point = ΔTb = kbm = (0.512°C/m)(0.690m) = 0.353°C
Boiling point of solution = (boiling point water) + ΔTb = 99.725°C + 0.353°C = 100.079°C
14.4. A solution was produced by dissolving 3.75 g of a nonvolatile solute in 95 g of acetone. The boiling point of pure acetone was observed to be 55.95°C, while the boiling point of the solution was 56.50°C. If the kb for acetone is 1.71°C/m, what is the approximate molar mass of the solute?
This problem is similar to Problem 14.2 except that the solvent is acetone, not water (disassociation does not occur in organic solvents; i = 1). Let us use the same technique as following Problem 14.2.
First compute the molality from the boiling-point equation.
Then, from the definition of molality, compute the number of moles solute, n(solute), in the sample.
And since
we solve for M:
14.5. The vapor pressure of water at 28°C is 28.35 torr. What is the vapor pressure at 28°C of a solution containing 68 g cane sugar, C12H22O11, in 1000 g of water?
Second Method
14.6. At 30°C, benzene (M = 78.1) has a vapor pressure of 121.8 torr. Dissolving 15.0 g of a nonvolatile solute in 250 g of benzene produced a solution with a vapor pressure of 120.2 torr. Determine the approximate molar mass of the solute.
Substituting in the relation v.p. solution = (v.p. pure solvent) × (mole fraction solvent),
Solving for M, we find the mass to be 350 g/mol. Note that the accuracy of the calculation is limited by the term 121.8 – 120.2 that appears in the expansion. The answer is significant to only 1 part in 16.
14.7. At 20°C the vapor pressure of methanol (CH3OH) is 94 torr and the vapor pressure of ethanol (C2H5OH) is 44 torr. Being closely related, these two alcohols form a two-component system which behaves pretty much as Raoult’s law predicts throughout the entire range of concentrations. A solution produced from 20 g CH3OH is mixed with 100 g C2H5OH. (a) Determine the partial pressure of each of the alcohols and the total vapor pressure of the solution. (b) Calculate the composition of the vapor above the solution by applying Dalton’s law (Chapter 5).
(a) In an ideal solution of two liquids, there is no distinction between solute and solvent. Raoult’s law holds for each of the components of ideal solutions. When two liquids are mixed to produce an ideal solution, the partial pressure of each liquid is equal to its partial pressure exerted by the pure substance multiplied by its mole fraction in solution. The molar mass of CH3OH is 32 and that of C2H5OH is 46.
The total pressure of the gaseous mixture is the sum of the partial pressures of all the components (Dalton’s law). The total pressure is 55 torr (21 + 34).
(b) Dalton’s law indicates that the mole fraction of any component of a gaseous mixture is equal to its pressure fraction, i.e., its partial pressure divided by the total pressure.
Since the mole fraction of (ideal) gases is the same as the volume fraction, we may also say that the vapor consists of 38% CH3OH by volume. Note that the vapor is relatively richer in the more volatile component, methyl alcohol (mole fraction 0.38), than is the liquid (mole fraction 0.22).
14.8. What would be the osmotic pressure at 17°C of an aqueous solution containing 1.75 g of sucrose (C12H22O11, also called cane sugar) per 150 mL of solution?
14.9. The osmotic pressure of a solution of a synthetic polyisobutylene in benzene was determined at 25°C. A sample containing 0.20 g of solute per 100 mL of solution developed a rise of 2.4 mm at osmotic equilibrium. The density of the solution was 0.88 g/mL. What is the molar mass of the polyisobutylene?
The osmotic pressure is equal to that of a column of the solution 2.4 mm high. By the formula in Chapter 5,
The molar concentration can now be determined from the osmotic pressure equation.
The solution contained 0.20 g solute per 100 mL solution, or 2.0 g/L, and has been found to contain 8.3 × 10–6 mol/L. Then,
14.10. An aqueous solution of urea had a freezing point of –0.52°C. Predict the osmotic pressure of the same solution at 37°C. Assume that the molar concentration and molality are numerically equal.
We can determine the molality by looking at the freezing-point lowering, 0.52°C.
The assumption that the molality and molarity are equal does not introduce serious error into the calculations for dilute aqueous solutions. The relationships discussed in Chapter 12 show that M ≈ m when the density is 1 g/mL (1 g/cm3) and M < 1000/m. Urea has a molar mass of 60.0 g/mol. Then, 0.280 mol/L may be used for the molarity in the osmotic pressure equation.
14.11. At 20°C and a total pressure of 760 torr, 1 L of water dissolves 0.043 g pure oxygen or 0.019 g pure nitrogen. Assuming that dry air is composed of 20% oxygen and 80% nitrogen, both by volume, determine the mass of oxygen and nitrogen dissolved by 1 L of water at 20° C and exposed to air at a total pressure of 760 torr.
The solubility of a gas (i.e., the concentration of dissolved gas) may be expressed as
This tells us that the solubility of a gas dissolved from a gaseous mixture (air in this problem) is directly proportional to the partial pressure of the gas. The proportionality constant, kH, is called Henry’s law constant. (Note: Some authors define the Henry’s law constant as the reciprocal of the kH used here.) To evaluate kH from the data, note that when pure oxygen is in equilibrium with water at a total pressure of 760 torr,
Then, allowing v.p. to be the vapor pressure of water, from the data,
When water is exposed to air at a total pressure of 760 torr,
Therefore,
Using the same technique, the solubility of N2 from air is (0.80)(0.019 g/L) = 0.015 g/L
14.12. A gaseous mixture of H2 and O2 contains 70% H2 and 30% O2 by volume. If the gas mixture is at 2.5 atm (excluding the vapor pressure of water) and is allowed to saturate water at 20°C, the water is found to contain 31.5 cm3 (S.T.P.) of H2/L. Find the solubility of H2 (reduced to S.T.P.) at 20°C and 1 atm partial pressure H2.
Since the volume of a gas at S.T.P. is proportional to its mass, the volume of the dissolved gas (reduced to S.T.P.) is proportional to the partial pressure of that gas.
14.13. A 25-mL sample of an aqueous solution containing 2 mg iodine is shaken with 5 mL of CCl4, then is allowed to separate (CCl4 and water do not mix). The solubility of iodine per unit volume is 85 times greater in CCl4 than in water at the temperature of the experiment and both saturated solutions may be considered to be “dilute.” (a) Calculate the quantity of iodine remaining in the water layer. (b) If a second extraction is made of the water layer using another 5 mL of CCl4, calculate the quantity of iodine remaining after this extraction.
The concentration of iodine in the water will be w mg iodine/25 mL water. The concentration of iodine in the CCl4 layer will be (2 – w)/5 (mg per mL CCl4). The relationship becomes
Note: Although this problem is given to us and worked in mg/mL, any volumetric concentration units could have been used in this problem, so long as the same units are used throughout because the conversion factors cancel.
(b) 
The concentration of iodine in the water layer will be y/25 and the concentration in the CCl4 layer will be (0.11 –y)/5. This second relationship becomes
14.14. A solution containing 6.35 g of a nonelectrolyte dissolved in 500 g of water freezes at –0.465°C. Determine the molar mass of the solute.
Ans. 50.8 g/mol
14.15. A solution containing 3.24 g of a nonvolatile nonelectrolyte and 200 g of water boils at 100.130°C at 1 atm. What is the molar mass of the solute?
Ans. 63.9 g/mol
14.16. Calculate the freezing point and the boiling point at 1 atm of a solution containing 30.0 g cane sugar (molar mass 342) and 150 g water.
Ans. –1.09°C; 100.300°C
14.17. Suppose you were asked to make up a 1.000m solution of cane sugar, C12H22O11, using a kilogram of water. (a) What is the calculated freezing point of the solution? (b) Is this a practical solution to produce?
Ans. (a) –1.86°C; (b) No, this solution would require 342 g sugar (± lb) to be dissolved in ±1 L of water, which is not likely to occur anywhere near the freezing point of water.
14.18. A solution of gold(III) nitrate is to be used in electroplating; it has been produced by dissolving 12.75 g Au(NO3)3 in 500 mL H2O (density 1.000). What are the expected freezing point and boiling point?
Ans. Assuming that i = 4 (1 gold and 3 nitrate ions), a freezing point of –0.495°C and boiling point of 100.14° C are expected.
14.19. If glycerin, C3H5(OH)3, and methanol, CH3OH, sold at the same price per pound, which would be cheaper for preparing an antifreeze solution for the radiator of an automobile?
Ans. Methanol (methyl alcohol). Since the molecular mass of methanol is much less than that of glycerin, a pound of methanol contains more molecules than a pound of glycerin. The amount of freezing-point depression is dependent on the number of particles (molecules here) in solution. This argument requires that the two substances be soluble in water, the other component in antifreeze.
14.20. How much ethanol, C2H5OH, must be added to 1 L of water so that the solution will not freeze above –4°F?
Ans. 495 g ethanol (ethyl alcohol)
14.21. A white powder is suspected of containing thallium(I) nitrate (molecular weight = 266). The freezing point of a solution (0.75 g powder dissolved in 50 g H2O) was –0.13°C. Assuming complete dissociation, could it be TlNO3?
Ans. No, the freezing point should have been –0.21°C.
14.22. What is the molecular mass of the white powder described in Problem 14.21?
Ans. 215 g/mol
14.23. A chemist must determine the molar mass of an unknown compound using a solvent that had a freezing point of 30.16°C. A solution of 0.617 g paradichlorobenzene, C6H4Cl2, in 10.00 g of solvent had a freezing point of 27.81°C, while 0.526 g of the unknown in 10.00 g of solvent lowered its freezing point to 26.47°C. Calculate the molar mass of the unknown.
Ans. 79.8 g/mol
14.24. Ethanol can be used for the short-term preservation of tissue. Calculate the freezing point of a solution mixed if you were to dissolve 0.500 g of a nonvolatile triose, C3H6O3, in 25 g ethanol (C2H5OH, freezing point pure = –114.6°C, kf = 1.99° C/m).
Ans. –115.0°C
14.25. What is the freezing point of a 10% (by weight) solution of CH3OH in water?
Ans. –6.5°C
14.26. A solution contains 10.6 g of a nonvolatile substance dissolved in 740 g of ether. The boiling point of the solution is 0.284°C over the boiling point of pure ether. Molal boiling-point constant for ether is 2.11°C · kg/mol. What is the molar mass of the substance?
Ans. 106 g/mol
14.27. The freezing point of a sample of naphthalene was found to be 80.6°C. When 0.512 g of a substance were dissolved in 7.03 g naphthalene, the solution’s freezing point was 75.2°C. What is the molar mass of the solute? (Molal freezing-point constant for naphthalene is 6.80°C · kg/mol.)
Ans. 92 g/mol
14.28. Pure benzene freezes at 5.45°C. A solution containing 7.24 g C2Cl4H2 in 115.3 g benzene was observed to freeze at 3.55°C. From these data, calculate the molal freezing-point constant for benzene.
Ans. 5.08°C · kg/mol
14.29. What is the osmotic pressure at 0°C of an aqueous solution containing 46.0 grams of glycerin (C3H5(OH)3, per liter?
Ans. 11.2 atm
14.30. An organic compound is known to be nonvolatile and a nonelectrolyte. A 0.35-g sample is dissolved in water and diluted to 150 mL. The osmotic pressure is measured as 0.04 atm at 25°C. What is the approximate mass number for this compound?
Ans. Approximately 1400 g/mol
14.31. From the data provided in Problem 14.30 and knowing that the density of the solution is 1.00 g/mL, (a) calculate the freezing point of the solution, and (b) determine if the freezing-point change would be a good way to determine the molecular mass of the compound. (c) Would the boiling-point change be a better determining factor than the freezing-point change?
Ans. (a) 0.0031°C; (b) The slight change in freezing point would be difficult to measure, making the freezing-point change not suitable for this determination. (c) The boiling-point change is only 0.00087°C and, therefore, even harder to measure with accuracy.
14.32. Salt is often used to melt ice and snow on roads in the winter and the temperature right now is –4.500°C. How much NaCl (complete dissociation) would be necessary to melt 1000 kg of ice by bringing down the freezing temperature to –4.500°C?
Ans. 70.7 kg
14.33. A solution of crab hemocyanin, a pigmented protein extracted from crabs, was prepared by dissolving 0.750 g in 125 mL of an aqueous medium. An osmotic pressure rise of 2.6 mm of the solution was detected at 4° C. The solution had a density of 1.00 g/mL. Determine the molar mass of the protein.
Ans. 5.4 × 105 g/mol (Note: Hemocyanin is to crabs as hemoglobin is to humans.)
14.34. The osmotic pressure of blood is 7.65 atm at 37°C. How much glucose, C6H12O6, should be used per liter for an intravenous injection to have the same osmotic pressure as blood?
Ans. 54.2 g/L
14.35. The vapor pressure of water at 26°C is 25.21 torr. What is the vapor pressure of a solution that contains 20.0 g glucose in 70 g water?
Ans. 24.51 torr
14.36. The vapor pressure of water at 25°C is 23.76 torr. The vapor pressure of a solution containing 5.40 g of a nonvolatile substance in 90 g water is 23.32 torr. What is the molar mass of the solute?
Ans. 57 g/mol
14.37. Ethylene bromide, C2H4Br2, and 1,2-dibromopropane, C3H6Br2, form ideal solutions over the entire range of composition. At 85°C the vapor pressures of the pure liquids are 173 torr and 127 torr, respectively. (a) Calculate the partial pressure of each component and the total pressure of the solution at 85°C if 10.0 g of ethylene bromide are dissolved in 80.0 g of 1,2-dibromopropane. (b) Calculate the mole fraction of ethylene bromide in the vapor in equilibrium with the above solution. (c) What would be the mole fraction of ethylene bromide in a solution at 85°C of a 50:50 mole mixture in the vapor?
Ans. (a) ethylene bromide, 20.5 torr; 1,2-dibromopropane, 112 torr; total, 132 torr; (b) 0.155; (c) 0.42
14.38. At 25°C, ethanol (C2H5OH) has a vapor pressure of 63 mm Hg, while 2-propanol’s (C3H7OH) vapor pressure is 45 mm Hg. What is the vapor pressure of a solution composed of 12 g ethanol and 27 g 2-propanol?
Ans. 53 mm Hg (53 torr)
14.39. An organic compound was found by combustion analysis to contain 38.7% C, 9.7% H, and the remainder oxygen. In order to determine its molecular formula, a 1.00 g sample was added to 10.00 g of water. The freezing point of the solution was found to be –2.94°C. What is the compound’s molecular formula?
Ans. C2H6O2
14.40. (a) A 0.100 molal solution of NaClO3 freezes at –0.3433°C. What would you predict for the boiling point of this aqueous solution at 1 atm? (b) At 0.001 molal concentration of this same salt, the electrical interferences between the ions no longer exist because the ions are, on the average, too far apart from each other for interaction. Predict the freezing point of the more dilute solution.
Ans. (a) 100.095°C; (b) –0.0037°C
14.41. The molar mass of a newly synthesized organic compound was determined by the isothermal distillation method. In this procedure two solutions, each in an open calibrated vial, are placed side by side in a closed chamber. One of the solutions contained 9.3 mg of the new compound, the other 13.2 mg of azobenzene (molar mass 182). Both were dissolved in portions of the same solvent. The experiment was untouched for three days, during which solvent distilled from one vial into the other until the same partial pressure of solvent was reached in both vials. After this, there was no further net distillation of solvent. Neither of the solutes distilled at all. The solution containing the new compound occupied 1.72 mL and the azobenzene solution occupied 1.02 mL. What is the molar mass of the new compound? The mass of solvent in solution may be assumed to be proportional to the volume of the solution.
Ans. 76 g/mol
14.42. Calculate the freezing point of a solution of 3.46 g of a compound, X, in 160 g of benzene, C6H6. When another sample of X was vaporized, its density was found to be 3.27 g/L at 116°C and 773 torr. The freezing point of benzene is 5.45°C and kf is 5.12°C kg/mol.
Ans. 4.37°C
14.43. In a particular chemical process, it is necessary to monitor a stream of benzene, C6H6, which may be contaminated with toluene, C7H8. How finely calibrated should the thermometer scale be to detect 0.10% by weight of toluene in the benzene by measuring its freezing point? (Refer to Problem 14.42.)
Ans. The freezing point is lowered to 0.056°C. You would need calibrations every 0.1°C at least; 0.01°C would be better. Such thermometers are not uncommon.
14.44. An alternative to the method of Problem 14.43 might be to measure the vapor pressure of the benzene stream at 25°C. Compare the two methods. Benzene and toluene form ideal solutions. The vapor pressure at 25°C is approximately 95 torr for benzene and 30 torr for toluene.
Ans. The lowering of the total vapor pressure of the liquid by the addition of 0.10% toluene is about 0.06 torr (0.06 mm Hg). Such an observation would require a microscope to read! Worse yet, since the vapor pressure of benzene increases about 3 torr/°C, it would be extremely difficult to control the sample temperature adequately to avoid error (better than within 0.01°C).
14.45. A method of producing drinkable water from salt water is reverse osmosis, in which a pressure just in excess of the osmotic pressure is applied to the solution in order to reverse the flow of the solvent (H2O). In principle, what pressure would be required to produce pure water from sea water at 25°C? Assume that sea water has a density of 1.021 g/mL and can be considered equivalent to 3.00% by weight NaCl, which is 100% ionized. Express your answer in atm, kPa, and psi.
Ans. 25.6 atm; 2.59 × 103 kPa; 376 psi
14.46. At 20°C and 1.00 atm partial pressure of hydrogen, 18 cm3 H2 (measured at S.T.P.) dissolve in 1 L of water. If water at 20°C is exposed to a gaseous mixture having a total pressure of 1400 torr (dry air) and containing 68.5% H2 by volume, what volume of H2 (at S.T.P.) which will dissolve in l L water.
Ans. 23 cm3
14.47. A liter of CO2 gas at 15°C and 1.00 atm dissolves in a liter of water also at 15°C when the pressure of the CO2 is 1 atm. Calculate the molar concentration of CO2 in solution over which the partial pressure of CO2 is 150 torr, also at 15°C.
Ans. 0.0083 M
14.48. (a) The solubility of iodine per unit volume is 200 times greater in ether than in water at a particular temperature. If an aqueous solution of iodine, 30 mL in volume and containing 2.0 mg of iodine, is shaken with 30 mL of ether and the ether is allowed to separate, what quantity of iodine remains in the water layer? (b) What quantity of iodine remains in the water layer if only 3 mL of ether are used? (c) How much iodine is left in the water layer if the extraction in (b) is followed by a second extraction, again using 3 mL of ether? (d) Which method is more efficient, a single large washing or repeated small washings?
Ans. (a) 0.010 mg; (b) 0.095 mg; (c) 0.0045 mg; (d) repeated small washings
14.49. The ratio of solubility of stearic acid per unit volume of n-heptane to that in 97.5% acetic acid is 4.95. How many extractions of a 10-mL solution of stearic acid in 97.5% acetic acid with successive 10-mL portions of n-heptane are needed to reduce the residual stearic acid content of the acetic acid layer to less than 0.5% of its original value?
Ans. 3 portions
14.50. Penicillin can be purified by extraction. The distribution coefficient for penicillin G between isopropyl ether and an aqueous phosphate medium is 0.34 (lower solubility in ether). The corresponding ratio for penicillin F is 0.68. A preparation of penicillin G has penicillin F as a 10.0% impurity. (a) If an aqueous phosphate solution of this mixture is extracted with an equal volume of isopropyl ether, what will be the % recovery of the initial G in the residual aqueous-phase product after one extraction? What will be the % impurity of this product? (b) Calculate the same two quantities for the product remaining in the aqueous phase after a second extraction with an equal volume of ether.
Ans. (a) 75% recovery and 8.1% impurity; (b) 56% recovery and 6.6% impurity
