Chapter 13 Properties of Solutions

1. Chapter 13Properties of Solutions Chemistry, The Central Science, 10th edition Theodore L. Brown; H. Eugene LeMay, Jr.; and Bruce E. Bursten Troy Wood University of Buffalo Buffalo, NY  2006, Prentice Hall

2. From weakest to strongest, rank the following solutions in terms of solvent–solute interactions: NaCl in water, butane (C4H10) in benzene (C6H6), water in ethanol. NaCl in water < C4H10 in C6H6 < water in ethanol Water in ethanol < NaCl in water < C4H10 in C6H6 C4H10 in C6H6 < water in ethanol < NaCl in water

3. Correct Answer: NaCl in water < C4H10 in C6H6 < water in ethanol Water in ethanol < NaCl in water < C4H10 in C6H6 C4H10 in C6H6 < water in ethanol < NaCl in water 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.

4. At a certain temperature, the Henry’s law constant for N2 is 6.0  104M/atm. If N2 is present at 3.0 atm, what is the solubility of N2? 6.0  104M 1.8  103M 2.0  104M 5.0  105M

5. Correct Answer: 6.0  104M 1.8  103M 2.0  104M 5.0  105M Henry’s law, Sg = (6.0  104M/atm)(3.0 atm) Sg = 1.8  103M

6. Determine the mass percentage of hexane in a solution containing 11 g of butane in 110 g of hexane. 9.0 % 10. % 90.% 91 %

7. Correct Answer: 9.0 % 10. % 90.% 91 % mass of component in solution =  mass % of component 100 total mass of solution Thus, 110 g (110 g + 11 g)  100 = 91%

8. 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

9. Correct Answer: 7.2 ppm 1.8 ppm 18 ppm 72 ppm mass of component in solution =  6 ppm of component 10 total mass of solution 3.6 mg 0.0036 g =  = 6 10 18 ppm 200. g 200. g

10. What is the molality of 6.4 g of methanol (CH3OH) dissolved in 50. moles of water? 0.040 m 0.22 m 0.064 m 0.11 m

11. Correct Answer: 0.040 m 0.22 m 0.064 m 0.11 m

12. How many moles of solute are there in 240 g of a solution that is 5.0% glucose (C6H12O6) by mass? 0.033 moles 0.067 moles 0.10 moles 0.12 moles 0.20 moles

13. Correct Answer: 0.033 moles 0.067 moles 0.10 moles 0.12 moles 0.20 moles 5.0 % glucose means 5.0 g glucose/100 g solution (5.0 g glucose/100 g solution)(240 g solution) = 12 g (12 g glucose) (1 mol glucose/180 g glucose) = 0.067 moles glucose

14. 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

15. Correct Answer: 9.0 torr 10. torr 80. torr 81. torr 90. torr  = P P i i total Pi = XiPtotal Pi = (0.72 mol/[0.72 + 0.080 mol])(90.0 torr) Pi = (0.90)(90.0 torr) = 81. torr

16. 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 CaCl2 in ethanol? 77.2°C 79.6°C 80.8°C 82.1°C 83.3°C

17. Correct Answer: 77.2°C 79.6°C 80.8°C 82.1°C 83.3°C = T K m  b b The increase in boiling point is determined by the molality of total particles in the solution. Thus, a 1.0 m solution of CaCl2 contains 1.0 m Ca2+ 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.

18. At 300 K, the osmotic pressure of a solution is 0.246 atm. What is its concentration of the solute? 1.0 M 0.50 M 0.25 M 0.10 M

19. Correct Answer: ( ) 1.0 M 0.50 M 0.25 M 0.10 M n = = RT MRT  V M= /RT M= (0.246 atm)/[(0.0821 L-atm/mol-K)(300K)] M= (0.246 atm)/(0.2463 L-atm/mol) = 1.0 M

20. Which of the following is not an example of a colloid? Fog Smoke Paint Milk Carbonated water

21. Correct Answer: Fog Smoke Paint Milk Carbonated water Carbonated water is a solution; all the other substances in the list are excellent examples of colloids.