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Author Topic: Physical Properties of Solutions  (Read 14924 times)

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meme

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Physical Properties of Solutions
« on: March 27, 2004, 01:02:58 PM »

Hi.

It's me again. Sorry for coming back so soon. We're having a test on Monday and our (not-so-very-good) professor gave us about 200 questions to practice. I'm having trouble working some out...

1. Which has the greater molal concentration (molality)?
a. 1.0 m KNO3
b. 1.0 M KNO3
c. Both have same molality.
The answer's supposed to be (b), but how can you convert it without knowing the density?

2. If the solubility of O2 from the air is 2.67 x 10-4 at sea level and 25 degrees C, what is the solubility of oxygen at an elevation of 12,000 ft where the atmospheric pressure is 0.657 atm. Assume the temperature is 25 degrees, and that the mole fraction of oxygen in air is 0.209 at both 12,000 ft and at sea level.
I think you need to use Raoult's Law to figure this out, but I've tried and tried again and I can't figure it out. The answer's supposed to be 1.75 x 10-4 M

3. Which of the following aqueous solutions highest boiling point? Given Kb for water is 0.52 degrees C/m.
a. 0.2 m KCl
b. 0.2 m Na2SO4
c. 0.2 m Ca(NO3)2
d. A and B
e. B and C
answer: (e). Okay.. if the formula is delta T = Kbm, and the Kb is the same, I would think that the boiling point would depend on the molality.. but they're all the same, so I'm lost here. ???

4. Which of the following aqueous solutions have the highest osmotic pressure at 25 degrees?
a. 0.2 M KBr
b. 0.2 M methanol
c. 0.2 M Na2SO4
d. 0.2 M KCl
answer: (c). Same question as above. Osmotic pressure = (molarity)(rate constant)(temperature). Everything's the same... how can you have different pressures ???

5. Which of the following aqueous solutions has the lowest freezing point? Given Kf for water is 1.86 degrees C/m.
a. 0.18 m KCl
b. 0.15 m Na2SO4
c. 0.12 m Ca(NO3)2
d. pure water
e. 0.20 m C2H6O2
answer: (b). Shouldn't freezing point lower with greater molality?

6. Arrange the following solutions in order of increasing boiling points: 0.050 m Mg(NO3)2, 0.100 m ethanol; 0.090 m NaCl.
answer: ethanol < Mg(NO3)2 < NaCl. Shouldn't boiling point increase with greater molality?

These next 2 I've tried a million and one times:

7. Calculate the approximate freezing poing of a solution made from 21.0g NaCl and 100g of H2O. Kf of water is 1.86degrees C/m.
answer: -13.4 degrees. Here's what I did:
converted NaCl to moles: .34 g NaCl
divided it by kg water to get molality: 3.59
multiplied by Kf to get delta Tf= 6.68
Which means that the freezing point would be -6.68 (which is one of the choices). am I doing something wrong?


8. What is the osmotic pressure of a solution prepared from 13.7g of the electrolyte HCl and enough water to make .50L of solution at 18 degrees?
answer: 35.9 atm. Okay, I figured out the molarity of the solution (which = .75M) and multiplied by constant (.0821) and temperature in kelvin and got: 17.9, which is also one of the choices, but apparently not the right one.

9. At 10 degrees one volume of water dissolves in 3.10 volumes of chlorine gas, at 1 atm pressure. What is Henry's law constant?
answer: 0.13. Law constant = molar concentration/pressure ...that's all I know

10. How many liters of ethylene glycol antifreeze C6H6O2 would you add to your car radiator containing 15.0 L of water if you eneded to protect your engine to -17.8 degrees? The density of ethylene glycol is 1.1 g/mL. For water Kf = 1.86.
answer: 8.1 L. No idea how to do this one at all, I've tried all these different things but can't get it.

11. The term "proof" is defined as twice the percent by volume of pure ehtanol in solution. A solution that is 95% ehtanol is 190 proof. What is the molarity of ethanol in 92 proof ethanol/water solution? density of ethanol = 0.80 g/cm3 density of water = 1.0 g/cm3.
answer: 8.0 M. All I've been able to work out is that its 46% ethanol. :(
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jdurg

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Re:Physical Properties of Solutions
« Reply #1 on: March 27, 2004, 02:23:28 PM »

Okay, I'll do my best to help you out here.

1).  Molality is moles of solute per kilogram of solution.  The molality of choice (a) is easy since it's given as 1.0 M/Kg.  For choice (b), you're given the molarity which is moles of solute per liter of solution.  Instead of trying to convert the molarity to molality, you might be better off converting the molality to molarity, then going ahead and determining which is more concentrated from that point.  (I.E. you should be able to easily determine the molarity of choice (a) and then use the molarity you calculated to set up a conversion factor for molarity to molality.  With your conversion factor, you should be able to convert choice (b) into molality).

2).  Try looking into Henry's Law.  ;)

3).  In the equation you're using, the molality is not the molality of the compound, it's the molality of the ions.  So in the case of choice (a), it's only 0.4 molal in total ions, while choices (b) and (c) are 0.6 molal in total ions.  (Freezing point depression and boiling point depression depend on the number of ions floating around, not what they are.  That is why they use CaCl2 during the winter.  It provides more ions than NaCl does).

4).  Again, osmotic pressure depends upon number of particles floating around.

5).  Once again, it's all determined by the number of ions/particles floating around, and choice (b) is 0.45 molal in terms of dissolved particles.

6).  Once again, see previous question.

Questions 7, 8, and 10).  Try these again, but after keeping in mind that the molality in the equations you're using is for dissolved ions/particles.  (So Na2SO4 would be more molal than NaCl would be).  

9).  Henry's Law is:  Sgas = KH * Pgas.  Sgas is the solubility of the gaseous component, KH is the Henry's Law Constant, and Pgas is the pressure of the dissolved gas over the solution.  From the information you should be able to easily solve for KH.

11).  This one's pretty easy.  They tell you off the bat that the proof is two times the percentage by volume.  From 92 proof we know that the solution is 46% ethanol and 54% water.  To make this easy we'll do this calculation on a one liter solution.  In one liter, 460 mL would be ethanol and 540 mL would be water.  To calculate molarity, we need to know the mass of the ethanol so that we could then calculate the number of moles.  (And since our calculations are on a one liter solution, the number of moles is our molarity).  With the density of ethanol at 0.80 g/mL, we know that 460 mL weighs 368 grams.  One mole of ethanol weighs 46.068 g/mol, so we can figure out that we have, to two significant figures, an 8.0 M ethanol solution.

I hope this has helped.   ;D
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Donaldson Tan

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Re:Physical Properties of Solutions
« Reply #2 on: March 27, 2004, 02:34:53 PM »

1. Which has the greater molal concentration (molality)?
a. 1.0 m KNO3
b. 1.0 M KNO3
c. Both have same molality.
The answer's supposed to be (b), but how can you convert it without knowing the density?
Molarity refers to the molar concentration of a solute inside a solvent..
2. If the solubility of O2 from the air is 2.67 x 10-4 at sea level and 25 degrees C, what is the solubility of oxygen at an elevation of 12,000 ft where the atmospheric pressure is 0.657 atm. Assume the temperature is 25 degrees, and that the mole fraction of oxygen in air is 0.209 at both 12,000 ft and at sea level.
I think you need to use Raoult's Law to figure this out, but I've tried and tried again and I can't figure it out. The answer's supposed to be 1.75 x 10-4 M
Did u try Henry's Law, whereby the solubility of a gas is directly proportional to the pressure acting on the solvent?

11. The term "proof" is defined as twice the percent by volume of pure ehtanol in solution. A solution that is 95% ehtanol is 190 proof. What is the molarity of ethanol in 92 proof ethanol/water solution? density of ethanol = 0.80 g/cm3 density of water = 1.0 g/cm3.
answer: 8.0 M. All I've been able to work out is that its 46% ethanol. :(
With reference to a 93 proof ethanol/water solution,
mass of ethanol in 100cm3 solution = 46X0.80 = 36.8g
mole of ethanol in 100cm3 solution = 36.8/46 = 0.80moles
molarity of ethanol = 0.80 / (100/1000) = 8.0M
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meme

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Re:Physical Properties of Solutions
« Reply #3 on: March 27, 2004, 06:06:22 PM »

Ohh.. okay, see the reason why I didn't get the last one is because we weren't given the molecular formula of ethanol or the molar mass. I'm guessing the prof. just forgot to include it. Anyway, thank you guys so much for your help, I really appreciate it.

I still don't have a clue about #10 though.

I tried using Henry's law for #2 and 9, but it still doesn't work for me.

and.. I understand the difference between molality and molarity, but I've tried converting the molarity to molality and vice versa for #1, and I can't do it. The examples our book shows us say that "the density of the solution serves as a converstion factor between molality and molarity"

I feel extremely stupid, but I'd be really grateful if one of you could explain these problems to me.  :-[
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Donaldson Tan

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Re:Physical Properties of Solutions
« Reply #4 on: March 27, 2004, 06:22:44 PM »

1. Is there a solvent present in choice (a) ?

2. oxygen solubility at sea level (1atm) = 2.65X10^(-4)
    oxygen solubility at 12000ft (0.657atm) = (0.657/1)X2.65X10^(-4) = 1.75X10(-4)
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meme

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Re:Physical Properties of Solutions
« Reply #5 on: March 27, 2004, 06:57:24 PM »

so you don't need to do anything with the mole fraction in #2?
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Re:Physical Properties of Solutions
« Reply #6 on: March 29, 2004, 04:51:56 AM »

the mole fraction is irrelevant for (2)
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Re: Physical Properties of Solutions
« Reply #7 on: July 12, 2009, 08:40:18 PM »

you will find physical properties of components on http://www.processglobe.com/Home.aspx
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