The intermolecular forces between solute and solvent particles must be strong enough to compete with those between solute particles and those between solvent particles.
If an ionic salt is soluble in water, it is because the ion-dipole interactions are strong enough to overcome the lattice energy of the salt crystal.
7.
From weakest to strongest, rank the following solutions in terms of solvent – solute interactions: NaCl in water, butane (C 4 H 10 ) in benzene (C 6 H 6 ), water in ethanol.
NaCl in water < C 4 H 10 in C 6 H 6 < water in ethanol
Water in ethanol < NaCl in water < C 4 H 10 in C 6 H 6
C 4 H 10 in C 6 H 6 < water in ethanol < NaCl in water
8.
Correct Answer: Butane in benzene will have only weak dispersion force interactions. Water in ethanol will exhibit much stronger hydrogen-bonding interactions. However, NaCl in water will show ion – dipole interactions because NaCl will dissolve into ions.
NaCl in water < C 4 H 10 in C 6 H 6 < water in ethanol
Water in ethanol < NaCl in water < C 4 H 10 in C 6 H 6
C 4 H 10 in C 6 H 6 < water in ethanol < NaCl in water
So even though enthalpy may increase, the overall energy of the system can still decrease if the system becomes more disordered.
15.
Water vapor reacts with excess solid sodium sulfate to form the hydrated form of the salt. The chemical reaction is Does the entropy increase or decrease? SAMPLE EXERCISE 13.1 Assessing Entropy Change
16.
Answer: The entropy increases because each gas eventually becomes dispersed in twice the volume it originally occupied. The water vapor becomes less dispersed (more ordered). When a system becomes more ordered, its entropy is decreased . PRACTICE EXERCISE Does the entropy of the system increase or decrease when the stopcock is opened to allow mixing of the two gases in this apparatus?
Vitamin A is soluble in nonpolar compounds (like fats).
Vitamin C is soluble in water.
26.
Solution C 7 H 16 is a hydrocarbon, so it is molecular and nonpolar. Na 2 SO 4 , a compound containing a metal and nonmetals, is ionic; HCl, a diatomic molecule containing two nonmetals that differ in electronegativity, is polar; and I 2 , a diatomic molecule with atoms of equal electronegativity, is nonpolar. We would therefore predict that C 7 H 16 and I 2 would be more soluble in the nonpolar CCl 4 than in polar H 2 O, whereas water would be the better solvent for Na 2 SO 4 and HCl. SAMPLE EXERCISE 13.2 Predicting Solubility Patterns Predict whether each of the following substances is more likely to dissolve in carbon tetrachloride (CCl 4 ) or in water: C 7 H 16 , Na 2 SO 4 , HCl, and I 2 .
27.
Answer: C 5 H 12 < C 5 H 11 Cl < C 5 H 11 OH < C 5 H 10 (OH) 2 (in order of increasing polarity and hydrogen-bonding ability) SAMPLE EXERCISE 13.2 continued PRACTICE EXERCISE Arrange the following substances in order of increasing solubility in water :
32.
At a certain temperature, the Henry’s law constant for N 2 is 6.0 10 4 M /atm. If N 2 is present at 3.0 atm, what is the solubility of N 2 ?
6.0 10 4 M
1.8 10 3 M
2.0 10 4 M
5.0 10 5 M
33.
Correct Answer: Henry’s law, S g = kP g S g = ( 6.0 10 4 M /atm)(3.0 atm) S g = 1.8 10 3 M
6.0 10 4 M
1.8 10 3 M
2.0 10 4 M
5.0 10 5 M
34.
SAMPLE EXERCISE 13.3 A Henry’s Law Calculation Calculate the concentration of CO 2 in a soft drink that is bottled with a partial pressure of CO 2 of 4.0 atm over the liquid at 25°C. The Henry’s law constant for CO 2 in water at this temperature is 3.1 10 –2 mol/L-atm. Solve: Check: The units are correct for solubility, and the answer has two significant figures consistent with both the partial pressure of CO 2 and the value of Henry’s constant. 2
10 6 Parts per Million (ppm) Parts per Billion (ppb) ppb = 10 9 mass of A in solution total mass of solution mass of A in solution total mass of solution
42.
If 3.6 mg of Na + is detected in a 200. g sample of water from Lake Erie, what is its concentration in ppm?
7.2 ppm
1.8 ppm
18 ppm
72 ppm
43.
Correct Answer: 6 10 solution of mass total solution in component of mass component of ppm ppm 18 10 g 200. g 0.0036 g 200. mg 3.6 6
If we know the density of the solution, we can calculate the molality from the molarity, and vice versa.
50.
SAMPLE EXERCISE 13.5 Calculation of Molality A solution is made by dissolving 4.35 g glucose (C 6 H 12 O 6 ) in 25.0 mL of water at 25°C. Calculate the molality of glucose in the solution. Solution molar mass of glucose, 180.2 g/mol water has a density of 1.00 g/mL, so the mass of the solvent is
Because of solute-solvent intermolecular attraction, higher concentrations of nonvolatile solutes make it harder for solvent to escape to the vapor phase.
P A is the normal vapor pressure of A at that temperature
NOTE: This is one of those times when you want to make sure you have the vapor pressure of the solvent .
56.
At a certain temperature, water has a vapor pressure of 90.0 torr. Calculate the vapor pressure of a water solution containing 0.080 mole sucrose and 0.72 mole water.
9.0 torr
10. torr
80. torr
81. torr
90. torr
57.
Correct Answer: total i P P i P i = X i P total P i = (0.72 mol/[0.72 + 0.080 mol])(90.0 torr) P i = (0.90)(90.0 torr) = 81. torr
9.0 torr
10. torr
80. torr
81. torr
90. torr
58.
Boiling Point Elevation and Freezing Point Depression
Nonvolatile solute-solvent interactions also cause solutions to have higher boiling points and lower freezing points than the pure solvent.
The change in boiling point is proportional to the molality of the solution:
T b = K b m
where K b is the molal boiling point elevation constant, a property of the solvent.
T b is added to the normal boiling point of the solvent.
60.
Ethanol normally boils at 78.4 ° C. The boiling point elevation constant for ethanol is 1.22 ° C/ m . What is the boiling point of a 1.0 m solution of CaCl 2 in ethanol?
77.2 ° C
79.6 ° C
80.8 ° C
82.1 ° C
83.3 ° C
61.
Correct Answer: The increase in boiling point is determined by the molality of total particles in the solution. Thus, a 1.0 m solution of CaCl 2 contains 1.0 m Ca 2+ and 2.0 m Cl for a total of 3.0 m . Thus, the boiling point is elevated 3.7 °C, so it is 78.4°C + 3.7°C = 82.1°C. m K T b b
Since these properties depend on the number of particles dissolved, solutions of electrolytes (which dissociate in solution) should show greater changes than those of nonelectrolytes.
We modify the previous equations by multiplying by the van’t Hoff factor, i
T f = K f m i
70.
PRACTICE EXERCISE Which of the following solutes will produce the largest increase in boiling point upon addition to 1 kg of water: 1 mol of Co(NO 3 ) 2 , 2 mol of KCl, 3 mol of ethylene glycol (C 2 H 6 O 2 )? Answer: 2 mol of KCl because it contains the highest concentration of particles, 2 m K + and 2 m Cl – , giving 4 m in all
In osmosis, there is net movement of solvent from the area of higher solvent concentration ( lower solute concentration ) to the are of lower solvent concentration ( higher solute concentration ).
The pressure required to stop osmosis, known as osmotic pressure , , is
where M is the molarity of the solution If the osmotic pressure is the same on both sides of a membrane (i.e., the concentrations are the same), the solutions are isotonic . n V = ( ) RT = MRT
If the solute concentration outside the cell is less than that inside the cell, the solution is hypotonic .
Water will flow into the cell, and hemolysis results.
79.
PRACTICE EXERCISE What is the osmotic pressure at 20°C of a 0.0020 M sucrose (C 12 H 22 O 11 ) solution? Answer: 0.048 atm, or 37 torr SAMPLE EXERCISE 13.11 Calculations Involving Osmotic Pressure The average osmotic pressure of blood is 7.7 atm at 25°C. What concentration of glucose (C 6 H 12 O 6 ) will be isotonic with blood? Solution Plan: Given the osmotic pressure and temperature, we can solve for the concentration, using Equation 13.13. Solve:
We can use the effects of a colligative property such as osmotic pressure to determine the molar mass of a compound.
81.
Solution Plan: K b for the solvent (CCl 4 )= 5.02ºC/ m . T b = K b m . SAMPLE EXERCISE 13.12 Molar Mass from Freezing-Point Depression A solution of an unknown nonvolatile electrolyte was prepared by dissolving 0.250 g of the substance in 40.0 g of CCl 4 . The boiling point of the resultant solution was 0.357°C higher than that of the pure solvent. Calculate the molar mass of the solute. Solve: The solution contains 0.0711 mol of solute per kilogram of solvent. The solution was prepared using 40.0 g = 0.0400 kg of solvent (CCl 4 ). The number of moles of solute in the solution is therefore
82.
PRACTICE EXERCISE Camphor (C 10 H 16 O) melts at 179.8°C, and it has a particularly large freezing-point-depression constant, K f = 40.0ºC/ m . When 0.186 g of an organic substance of unknown molar mass is dissolved in 22.01 g of liquid camphor, the freezing point of the mixture is found to be 176.7°C. What is the molar mass of the solute? Answer: 110 g/mol
83.
SAMPLE EXERCISE 13.13 Molar Mass from Osmotic Pressure The osmotic pressure of an aqueous solution of a certain protein was measured in order to determine the protein’s molar mass. The solution contained 3.50 mg of protein dissolved in sufficient water to form 5.00 mL of solution. The osmotic pressure of the solution at 25°C was found to be 1.54 torr. Calculate the molar mass of the protein. Solution Plan: The temperature ( T = 25ºC) and osmotic pressure ( = 1.54 torr) are given, and we know the value of R so we can use Equation 13.13 to calculate the molarity of the solution, M . In doing so, we must convert temperature from °C to K and the osmotic pressure from torr to atm. We then use the molarity and the volume of the solution (5.00 mL) to determine the number of moles of solute. Finally, we obtain the molar mass by dividing the mass of the solute (3.50 mg) by the number of moles of solute. Solve: Solving Equation 13.13 for molarity gives Because the volume of the solution is 5.00 ml = 5.00 10 –3 L, the number of moles of protein must be
84.
Comment: Because small pressures can be measured easily and accurately, osmotic pressure measurements provide a useful way to determine the molar masses of large molecules. PRACTICE EXERCISE A sample of 2.05 g of polystyrene of uniform polymer chain length was dissolved in enough toluene to form 0.100 L of solution. The osmotic pressure of this solution was found to be 1.21 kPa at 25°C. Calculate the molar mass of the polystyrene. Answer: 4.20 10 4 g/mol SAMPLE EXERCISE 13.13 continued The molar mass is the number of grams per mole of the substance. The sample has a mass of 3.50 mg = 3.50 10 –3 g. The molar mass is the number of grams divided by the number of moles:
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