CHE-122 Written Assignment 2 PDF

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Name: College ID: Thomas Edison State University General Chemistry II with Labs (CHE-122) Section no.: OL009 Semester and year: JUN2018

Written Assignment 2: Physical Properties of Solutions Answer all assigned questions and problems, and show all work. 1. Arrange the following compounds in order of increasing solubility in water: O2, LiCl, Br2, CH3OH (methanol) (4 points)

1. O2 2. LiCl 3. Br2 4. CH3OH (Reference: Chang 12.11)

2. Calculate the percent by mass of the solute in each of the following aqueous solutions: (12 points) a. 5.50 g of NaBr in 78.2 g of solution 5.5g / 78.2g= 7% b. 31.0 g of KCl in 152 g of water 31g/(31g+152g)=16.9% c. 4.5 g of toluene in 29 g of benzene 4.5g / (4.5g+29g)= 13.4% (Reference: Chang 12.15)

3. Calculate the molality of each of the following solutions: (8 points) a. 14.3 g of sucrose (C12H22O11) in 676 g of water 14.3g / (342.3g/mol * 676g)= 0.0000618=0.0618mol/kg b. 7.20 moles of ethylene glycol (C2H6O2) in 3546 g of water 7.2mol / 3546g=0.00203mol/g=2.03mol/kg 1 Copyright © 2017 by Thomas Edison State University. All rights reserved.

(Reference: Chang 12.17)

4. Calculate the molalities of the following aqueous solutions: (12 points) a. 1.22 M sugar (C12H22O11) solution (density of solution = 1.12 g/mL) 1.22mol/L / (1000ml/L * 1.12g/ml - 342.3g/mol * 1.22mol/L) * 1000ml=1.737mol/kg b. 0.87 M NaOH solution (density of solution = 1.04 g/mL) 0.87mol/L / (1000ml/L * 1.04g/ml – 40g/mol * 0.87mol/L) * 1000ml/L=0.865mol/kg c. 5.24 M NaHCO3 solution (density of solution = 1.19 g/mL) 5.24mol/L / (1000ml/L * 1.19g/ml – 84g/mol * 5.24mol/L) *1000ml/L= 7mol/kg (Reference: Chang 12.19)

5. The alcohol content of hard liquor is normally given in terms of the “proof,” which is defined as twice the percentage by volume of ethanol (CH3OH) present. Calculate the number of grams of alcohol present in 1.00 L of 75-proof gin. The density of ethanol is 0.798 g/mL. (4 points) 75proof= 37.5% 1L *37.5%= 375ml of the bottle is ethanol. 375ml * 0.798g/ml=299.25g of ethanol (Reference: Chang 12.21)

6. Calculate the molarity and the molality of an NH3 solution made up of 30.0 g of NH3 in 70.0 g of water. The density of the solution is 0.982 g/mL. (8 points) 30g/ 17g/mol=1.76mol 1.76mol / 0.102L= 17.3M

100g/0.982g/ml=102ml 1.76mol/70g=25.2mol/kg (Reference: Chang 12.23) 2

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7. Concentrated nitric acid (HNO3) has a density of 1.42 g/mL and is 70.0% nitric acid. Calculate the molarity of the acid if 30.0 mL of the concentrated acid is mixed with 255 mL of water. (5 points) 1.42g/ml * 30ml=42.6g 29.8g / 63g/mol=0.473mol 0.473mol/0.285L=1.66M

70% * 42.6g=29.8g 30ml + 255ml=0.285L

8. A 3.20 g sample of a salt dissolves in 9.10 g of water to give a saturated solution at 25°C. a. What is the solubility (in g salt/100 g of water) of the salt? (2 points) 3.2g / 9.1g= 0.3516 or 35.16% salt So if 100g of water is used, 35.16g of salt would be dissolved. b. How much water would it take to dissolve 25 g of this salt? (2 points) 25g / 0.3516= 71.1g c. If 10.0 g of this salt is mixed with 15.0 g of water, what percentage of the salt dissolves? (2 points) 0.3516 * 15g= 5.27g of salt out of the 10g total. 5.27 / 10= 52.7% (Reference: Chang 12.27)

9. A man bought a goldfish in a pet shop. Upon returning home, he put the goldfish in a bowl of recently boiled water that had been cooled quickly. A few minutes later the fish was found dead. Explain what happened to the fish. (3 points) The fish basically suffocated to death because all of the oxygen had been released from the water when it was boiled and the water had not been given the chance to absorbed/dissolved oxygen prior to the fish being placed in the tank. (Reference: Chang 12.34)

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10. What are the boiling point and freezing point of a 2.47 m solution of benzene in benzene? (The boiling point and freezing point of naphthalene are 80.1ºC and 5.5ºC, respectively.) Kb = 2.53 ºC /m ; Kf = 5.12 ºC /m (6 points) 2.53 ºC /m * 2.47m= 6.25ºC 80.1ºC – 6.25 ºC= 73.85 ºC 5.12 ºC /m * 2.47m= 12.65 ºC

5.5 ºC – 12.65 ºC= -7.15 ºC (Reference: Chang 12.55)

11. Pheromones are compounds secreted by the females of many insect species to attract males. One of these compounds contains 80.78 percent carbon, 13.56 percent hydrogen, and 5.66 percent oxygen. A solution of 1.00 g of this pheromone in 8.50 g of benzene freezes at 3.37ºC. What are the molecular formula and molar mass of the compound? (The normal freezing point of pure benzene is 5.50ºC.) Kf = 5.12 ºC /m (10 points) 5.5 ºC – 3.37 ºC=2.13 ºC 2.13ºC / 5.12ºC/m= 0.416mol/kg 80.78/12= 6.73 13.56/1=13.56 5.66/16=0.354 6.73 : 13.56 : 0.354 = *3 =C20H40O= (20*12) + (40*1) + (16*1)= 296g/mol (Reference: Chang 12.57)

12. How many liters of the antifreeze ethylene glycol [CH2(OH)CH2(OH)] would you add to a car radiator containing 6.50 L of water if the coldest winter temperature in your area is –20ºC? Calculate the boiling point of this water-ethylene glycol mixture. (Te density of ethylene glycol is 1.11 g/mL.) Kf = 1.86 ºC /m (8 points) deltaT=20ºC 20ºC / 1.86ºC/m= 10.75mol/kg 10.75mol/kg * 6.5kg=69.89mol 69.89mol * 62g/mol=4333g 4333g / 1.11g/ml=3903.6ml=3.9L 0.512ºC/m * 10.75m= 5.5ºC 100 ºC – 5.5 ºC= 94.5 ºC=boiling point. (Reference: Chang 12.59)

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13. What is the osmotic pressure (in atm) of a 1.36 M aqueous solution of urea [(NH2)2CO] at 22.0ºC? (3 points) 1.36M * 0.0821Latm / mol 295K=32.9atm osmotic pressure (Reference: Chang 12.63)

14. A quantity of 7.480 g of an organic compound is dissolved in water to make 300.0 mL of solution. The solution has an osmotic pressure of 1.43 atm at 27ºC. The analysis of this compound shows that it contains 41.8 percent C, 4.7 percent H, 37.3 percent O, and 16.3 percent N. Calculate the molecular formula of the compound. (6 points) C3H4O2 3*12 + 4+ 16*2= 72 429/72= 5.95 is roughly 6 C3H4O2 * 6= C18H24O12 which is 432g/mol (Reference: Chang 12.65)

15. Arrange the following solutions in order of decreasing freezing point: 0.10 m Na3PO4, 0.35 m NaCl, 0.20 m MgCl2, 0.15 m C6H12O6, 0.15 m CH3COOH. (5 points) 1. 0.15 m C6H12O6 2. 0.15 m CH3COOH 3. 0.10 m Na3PO4 4. 0.20 m MgCl2 5. 0.35 m NaCl (Reference: Chang 12.71)

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