EVEN (i.e.top) PROBLEMS:

1. 25 g. of NaCl are dissolved in enough water to make 1.0 liters solution. Assume the density is 1.0 g/cm³, and the molecular weight of NaCl is 58.44 g./mole. Calculate the molarity of NaCl.
   1. 0.43 M
   2. 0.39 M
   3. 0.46 M
   4. 4.6 M
   5. 0.52 M
   ANSWER: Molarity is just moles solute divided by liters of solution. Therefore, it is only necessary to divide 25 g. NaCl by 58.44 g./mole NaCl (because we are already talking about 1.0 liter of solution here) to obtain 0.43 M, A.

2. 32 g. of KCl are dissolved in enough water to make 1.0 liters solution. Assume the density is 1.0 g/cm³, and the molecular weight of KCl is 74.55 g./mole. Calculate the molality of KCl.
   1. 0.54 m
   2. 0.36 m
   3. 0.39 m
   4. 0.44 m
   5. 0.49 m
   ANSWER: Molality is moles solute divided by kg. solvent. How many kg. of water are present in 1.0 liters of this solution? Well, the density of the solution is 1.0 g./cm³ so the weight of water in 1.0 liter solution will be 1000 g. soln - 32 g. NaCl = 968 g. water or 0.968 kg. water. The molality is therefore: 32 g. NaCl/74.55 g./mole NaCl divided by 0.968 kg. water = 0.44 m, D.

3. Which of the following solutions will have the lowest TOTAL vapor pressure?
   1. Pure water.
   2. Water + 0.1 M ethyl ether.
   3. Water + 0.1 M NaCl
   4. Water + 0.1 M MgCl₂
   5. Pure ethyl ether.
   ANSWER: Ethyl ether is obviously volatile - in fact, its volatility is much higher than that of water. On the other hand, NaCl and MgCl₂ are both salts having zero volatility. According to Raoult`s Law, these involatile components will reduce the vapor pressure of water above a solution. And the mole fraction of water will be lower in the 0.1 M MgCl₂ solution than in 0.1 M NaCl solution because MgCl₂ liberates three ions in water, whereas NaCl liberates just two. The correct answer, then, must be D.

4. REVERSE OSMOSIS ...(which of the following is correct?)
   
   1. ...is a way to depress the freezing point of water.
   2. ...is a way to increase the boiling point of water.
   3. ...uses pressure to freeze water above 0^oC.
   4. ...uses pressure to purify impure water.
   5. None of the above.
   ANSWER: Well, statements A, B, and C are nonsense, of course. Statement D, however, is correct. HOWEVER because of an error in Problem 4 for the ODD folks (see below), everyone will receive full credit for this problem.

ODD (i.e. bottom) PROBLEMS:

1. 25 g. of NaCl are dissolved in enough water to make 1.0 liters solution. Assume the density is 1.0 g/cm³, and the molecular weight of NaCl is 58.44 g./mole. Calculate the molality of NaCl.
   1. 0.40 m
   2. 0.44 m
   3. 0.38 m
   4. 0.52 m
   5. 6.7 m
   ANSWER: Molality is moles solute divided by kg. solvent. How many kg. of water are present in 1.0 liters of this solution? Well, the density of the solution is 1.0 g./cm³ so the weight of water in 1.0 liter solution will be 1000 g. soln - 25 g. NaCl = 975 g. water or 0.975 kg. water. The molality is therefore: 25 g. NaCl/58.44 g./mole NaCl divided by 0.975 kg. water = 0.44 m, B.

2. 32 g. of KCl are dissolved in enough water to make 1.0 liters solution. Assume the density is 1.0 g/cm³, and the molecular weight of KCl is 74.55 g./mole. Calculate the molarity of KCl.
   1. 0.64
   2. 0.57
   3. 0.39
   4. 0.48
   5. 0.43
   ANSWER: Molarity is just moles solute divided by liters of solution. Therefore, it is only necessary to divide 32 g. NaCl by 74.55 g./mole NaCl (because we are already talking about 1.0 liter of solution here) to obtain 0.43 M, E.

3. Which of the following solutions will have the lowest TOTAL vapor pressure?
   1. Pure water.
   2. Water + 0.1 M CaF₂
   3. Water + 0.1 M NaF
   4. Water + 0.1 M dimethyl ketone
   5. Pure dimethyl ketone.
   ANSWER: Dimethyl ketone is obviously volatile - in fact, its volatility is higher than that of water. On the other hand, NaF and CaF₂ are both salts having zero volatility. According to Raoult`s Law, these involatile components will reduce the vapor pressure of water above a solution. And the mole fraction of water will be lower in the 0.1 M CaF₂ solution than in 0.1 M NaF solution because MgCl₂ liberates three ions in water, whereas NaF liberates just two. The correct answer, then, must be B.

4. At high pressures, water... (which of the following is correct?)
   
   1. Becomes less viscous.
   2. Has higher compressibility.
   3. Boils below 100^oC.
   4. Freezes above 0^oC.
   5. All of the above.
   ANSWER: We haven`t talked about the compressibility or the viscosity of water at high pressures, but in any case, these two statements - A & B - are false. At high pressures, the boiling point is HIGHER than at lower pressures, not lower so C is incorrect. And because the density of ice is less than that of water, it is MORE DIFFICULT to freeze water at high pressure so the freezing point would actually be LOWER and D is false. In other words, A-E are ALL incorrect. Sorry for the scare! Everyone will receive full credit for this problem.