Chapter 13 Solutions

Chapter 13Solutions

2.0 mM 1.0 Solubility 0 0 15 30 45 60 Temperature ( ° C) The curve describes the solubility properties of • A liquid • An ionic solid • A covalent solid • A gas • A gas or liquid

Considering the substances given below, predict which will be soluble in water. • 1 • 2 • 3 • 2, 3 • 1, 2, 3

When concentration is expressed in units of ________ , changing temperature will affect the solution concentration. • Molality • Molarity • ppm • Molarity and molality • Molarity, molality, and ppm

Predict which aqueous solution will have the lowest freezing point. • 0.25 m C2H5OH • 0.15 m CaCl2 • 0.20 m NaCl • 0.15 m NH4NO3 • 0.15 m Na3PO4

Arrange the aqueous solutions according to increasing boiling point. • AlCl3 < KNO3 < Na2SO4 • Na2SO4 < AlCl3 < KNO3 • Na2SO4 < KNO3 < AlCl3 • KNO3 < AlCl3 < Na2SO4 • KNO3 < Na2SO4 < AlCl3 0.10 m Na2SO4 0.15 m AlCl3 0.20 m KNO3

Which polyethylene solution will produce the largest osmotic pressure? Assume that all solutions use the same solvent and have the same concentration.

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

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.

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

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

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

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%

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

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

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

Correct Answer: 0.040 m 0.22 m 0.064 m 0.11 m

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

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

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

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

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

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.

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

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

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

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.

Chapter 13 Solutions