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Q: Lowering the freezing point of water ( Answered ,   7 Comments )
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 Subject: Lowering the freezing point of water Category: Science > Chemistry Asked by: brittanyl-ga List Price: \$10.00 Posted: 25 Jul 2005 06:21 PDT Expires: 24 Aug 2005 06:21 PDT Question ID: 547572
 ```Which lowers the freezing point of 2.0 Kg of water more, 0.20 mol or .20 mol of Ba(OH)2? Why?``` Clarification of Question by brittanyl-ga on 25 Jul 2005 13:30 PDT `That is suppose to be .20 moles of NaOh or .20 mol of Ba(OH)2? Sorry.`
 Subject: Re: Lowering the freezing point of water Answered By: welte-ga on 28 Jul 2005 16:40 PDT Rated:
 ```Hi Brittanyl-ga, The concept behind the problem you pose is known as Raoult's Law. This law can be posed in essentially two different ways: The lowering of the freezing point of a solvent is directly proportional to the number of solute molecules (or moles) in a given weight of solvent. The proportionality constant can be measured for each solvent. Another way to put it is as follows (from InfoPlease.com): "Quantitatively, Raoult's law states that the solvent's vapor pressure in solution is equal to its mole fraction times its vapor pressure as a pure liquid, from which it follows that the freezing point depression and boiling point elevation are directly proportional to the molality of the solute, although the constants of proportion are different in each case." ________________________ So, what does this all mean? Basically, adding solute (such as NaOH) to a solvent (such as water) causes the solute to "hold on" to solvent molecules that would otherwise escape from the liquid (in the case of boiling point raising) or bind to other like solvent molecules during the formation of the solid phase (in the case of freezing point reduction). The mathematical relation is relatively simple, but is only accurate for dilute solutions. More concentrated solutions become more complicated. You can read more about Raoult's law here: http://www.infoplease.com/ce6/sci/A0841143.html A good description of how to do calculations using this law can be found at Unit5.org: http://www.unit5.org/christjs/Molality_Freezing_Lowering.htm As stated at this site, it's important to recognize how many particles a solute dissociates into on solution. For example, in the case of ionic salts, such as NaCl, the solute will dissociate into Na+ and Cl-, two particles. In the case of NaOH, we have Na+ and OH-. For Ba(OH)2, we have Ba+ and 2 ions of OH-, three particles. We must multiple the molality of our solute by the number of particles liberated upon formation of the solution to get the right answer. That is the key to your problem. For NaOH and Ba(OH)2, because the solvet and molar concentrations of the solvents are the same, we can see that Ba(OH)2 will lower the freezing point (same as the melting point) more, because it liberates 3 particles on solution. If you like, we can actually calculate the freezing point of each of the solutions, although it's useful to develop the skill to see this answer based on the formula without having to resort to the actual calculations. ________________________ Here we go... For NaOH: The basic formula for freezing point depression is Freezing point = FP(without solute) - DT_f (depression of freezing temperature) All of the values need to be in degrees Celsius or Kelvin. FP = 0 deg Celsius for water, as I'm sure you know. The formula for DT_f is DT_f = k_f x m x N, where N is the number of particles we get on solution, as I discussed above; m is the molality of the solution, k_f is a constant that depends on the solvent. k_f = 1.86 deg C/mol for water. so, DT_f = 1.85 deg C/mol x 0.20 mol x 2 = 0.74 deg C Similarly for Ba(OH)2, DT_f = 1.85 deg C/mol x 0.20 mol x 3 = 1.11 deg C So, for 0.2 molar NaOH, the freezing point will be -0.74 deg Celsius. For Ba(OH)2, the freezing point will be -1.11 deg Celsius. ________________________ Also, as an aside, one commenter mentioned that salt used on roads in the winter is typically KCl. In most states, the most common road salt is actually CaCl2. As you can see, compared to NaCl, CaCl2 gives us 3 particles in solution instead of 2 from NaCl. A problem, however, is that CaCl2 is corrosive and somewhat toxic. Also, in this application, the solute is at a very high concentration, which violates the assumption I mentioned above (that the solution is dilute). If you're interested, you can read more about winter road salt here: http://www.usroads.com/journals/p/rmj/9712/rm971202.htm A nice comparison of road salts can be found here: http://www.city.vancouver.bc.ca/ctyclerk/cclerk/980407/a5.htm ________________________ I hope this was helpful. Best of luck in your studies. -welte-ga```
 brittanyl-ga rated this answer: `Very helpful and very descriptive answer!`

 Subject: Re: Lowering the freezing point of water From: pafalafa-ga on 25 Jul 2005 06:58 PDT
 `.2 mol, definitely.`
 Subject: Re: Lowering the freezing point of water From: brittanyl-ga on 25 Jul 2005 07:45 PDT
 `.2 mol of what?`
 Subject: Re: Lowering the freezing point of water From: stephanbird-ga on 26 Jul 2005 01:17 PDT
 `Why should either of them lower the freezing point at all?`
 Subject: Re: Lowering the freezing point of water From: hfshaw-ga on 26 Jul 2005 15:52 PDT
 ```Sounds like a homework problem. The formula for freezing point depression is : dT = i*K*m, where K is a constant specific to the solvent in question (equal to -1.858 Kelvins/molal for water); m is the molality of the solution; and i is the "van't Hoff factor", a unitless number that indicates the degree of dissociation of the dissolved solute. It is equal to 1 for solvents that do not dissociate, and equal to the number of moles of ions produced per mole of solute for solvents that do dissociate. NaOH dissociates into 2 moles of ions per mole of solute, while Ba(OH)2 dissociates into 3 moles of ions per mole of solute. From this, you should be able to say which solute will depress the freezing point of water more.```
 Subject: Re: Lowering the freezing point of water From: greyw01f-ga on 27 Jul 2005 07:55 PDT
 ```Answer: BaOH2 lowers the freezing point of water more, and here's why: When water freezes, what happens is that the individual water molecules organize themselves into very orderly crystals. When there's a foreign material (ie. BaOH2 or NaOH) in the mix, it disrupts the "orderliness" of the crystals. It therefore requires colder temperatures to form ice crystals, thereby lowering the freezing point. When either compound (BaOH2 or NaOH) is added to water, the ions dissociate completely (because they're both strong bases). Not only (as hfshaw-ga correctly stated) are there more ions in the Ba solution (two OH-'s instead of one with NaOH), but the Ba++ ion is much larger and more positively charged than the Na+ ion. This causes more disruption of the crystallization process, making it harder for ice crystals to form. This is exactly why the expensive "road salt" that you buy in stores is much better at removing ice than "rock salt." "Rock salt" is essentially NaCl, and the more expensive white "road salt" is primarily KCl. Potassium is much larger than sodium and, even though they have the same charge (+1), the larger potassium ion disrupts the orderly ice crystal formation, thereby lowering the freezing temperature of water.```
 Subject: Re: Lowering the freezing point of water From: njhvactech-ga on 28 Jul 2005 20:28 PDT
 ```I know for a fact that sugar lowers the freezing point of water. An Icee/slurpee has a freezing point of 26 degrees, that's why you get brain freeze when you drink it.```
 Subject: Re: Lowering the freezing point of water From: nice360t-ga on 08 Aug 2005 19:40 PDT
 ```It should be mentioned that the answer depends on the solubility of the salts involved. Barium hydroxide has limited solubility, .2 moles in 2 kg water is just under the limit for cool water. For slightly different conditions (e.g., less water) the answer would be the sodium hydroxide, since it is much more soluble. A better example to show the principle would be to compare Sodium sulfate (Na2SO4), which produces three ions and is very soluble.```