DivideByZero Posted March 6, 2008 Share Posted March 6, 2008 using only water, ice and one of the substances below, im suppose to lower the temperature of water as much as possible. I was given 10 substances and I have successfully lowered them down to 2 choices. So the substances left for me to choose are: 1. CaCl2 • 2H20 2. Na2Co3 Which substance should I mix with about 10mL of water+ice for it to reach a very low temperature? I know that the more ions the substance has the lower the water's temperature will become. But right now I'm confused between CaCl2•2H20 and Na2Co3. Any help please? Link to comment Share on other sites More sharing options...
swansont Posted March 6, 2008 Share Posted March 6, 2008 Two things will happen: freezing point depression and a endo- or exothermic reaction when the material dissolves. Freezing point depression is a colligative property. That is, it depends on the number of moles of solute. Remembering that salts disassociate into their constituent ions when they dissolve, which will yield the better result? Link to comment Share on other sites More sharing options...
DivideByZero Posted March 6, 2008 Author Share Posted March 6, 2008 Two things will happen: freezing point depression and a endo- or exothermic reaction when the material dissolves. Freezing point depression is a colligative property. That is, it depends on the number of moles of solute. Remembering that salts disassociate into their constituent ions when they dissolve, which will yield the better result? Ah thats interesting! 1 mole of CaCl2 = 110.984g 1 mole of Na2Co3 = 105.987g 1 mole of MgSO4 = 120.366g if I am only given 20g of a solution I have: 20g of CaCl2 = 0.18 moles 20g of Na2Co3 = 0.19 moles 20g of MgSO4 = 0.17 moles Given this data I found that the freezing point of Na2CO3 is -106.02 freezing point of CaCl2 is -100.44 and freezing point of MgSO4 is -61.75 So Am I correct? Will Na2CO3 produce the lowest temperature? Please inform me ASAP (before tomorrow)! By the way, thanks a lot ! Link to comment Share on other sites More sharing options...
swansont Posted March 6, 2008 Share Posted March 6, 2008 Ah thats interesting!1 mole of CaCl2 = 110.984g 1 mole of Na2Co3 = 105.987g 1 mole of MgSO4 = 88.3g if I am only given 20g of a solution I have: 20g of CaCl2 = 0.18 moles 20g of Na2Co3 = 0.19 moles 20g of MgSO4 = 0.23 moles Therefore (out of the three) MgSO4 should have yield the lowest freezing point! Am I correct? (note this also works out when comparing density) CaCl2 is a salt, so for each mole of the solid, you get a mole of Ca and 2 moles of Cl in solution. So for that, it's .54 moles of solute in 20g. Figure out what the dissolved products are for the others. Link to comment Share on other sites More sharing options...
thedarkshade Posted March 6, 2008 Share Posted March 6, 2008 So for that, it's .54 moles of solute in 20g. Actually in 20 g of [ce]CaCl2[/ce] you got 0.18 moles. [math]n=\frac{m}{M}=\frac{20g}{110\frac{g}{mol}}=0.181818mol[/math] Link to comment Share on other sites More sharing options...
DivideByZero Posted March 6, 2008 Author Share Posted March 6, 2008 solute = CaCl2 molarity = 20g/10mL = 0.18moles/0.010L = 18M produces 3 moles of particles 3*18M = 54M KfM = (1.86)(54M) = 100.44 Freezing point = -100.44 solute = Na2CO3 molarity = 20g/10mL = 0.19moles/0.010L = 19M produces 3 moles of particles 3*19M = 57M KfM = (1.86)(57M) = 106.02 Freezing point = -106.02 solute = MgSO4 molarity = 20g/10mL = 0.166moles/0.010L = 16.6M produces 2 particles 2*16.6M = 33.2M KfM = (1.86)(33.2M) = 61.75 Freezing point = -61.75 Therefore Na2CO3 has the lowest freezing point!!!!!!!! correct?!? please reply!! Link to comment Share on other sites More sharing options...
swansont Posted March 7, 2008 Share Posted March 7, 2008 Actually in 20 g of [ce]CaCl2[/ce] you got 0.18 moles. [math]n=\frac{m}{M}=\frac{20g}{110\frac{g}{mol}}=0.181818mol[/math] 0.18 moles of [ce]CaCl2[/ce]. But when it dissolves, you have 0.18 moles of Ca++ and 2 x 0.18 moles of Cl- = 0.54 moles of dissolved ions, and that's what the freezing point depression depends on. solute = CaCl2molarity = 20g/10mL = 0.18moles/0.010L = 18M produces 3 moles of particles 3*18M = 54M KfM = (1.86)(54M) = 100.44 Freezing point = -100.44 solute = Na2CO3 molarity = 20g/10mL = 0.19moles/0.010L = 19M produces 3 moles of particles 3*19M = 57M KfM = (1.86)(57M) = 106.02 Freezing point = -106.02 solute = MgSO4 molarity = 20g/10mL = 0.166moles/0.010L = 16.6M produces 2 particles 2*16.6M = 33.2M KfM = (1.86)(33.2M) = 61.75 Freezing point = -61.75 Therefore Na2CO3 has the lowest freezing point!!!!!!!! correct?!? please reply!! While the calculations look OK — I think you have the right compound — but the freezing point numbers don't look quite right. Did you check the solubility to make sure you can actually dissolve that much? As I recall the freezing point depression for [ce]CaCl2[/ce] is around -20 ºF, which is around -30 ºC http://chemistry.about.com/cs/howthingswork/a/aa120703a.htm Though the website says the lowest practical temperature, so that may be limitations other than solubility that might not apply here. Link to comment Share on other sites More sharing options...
DivideByZero Posted March 7, 2008 Author Share Posted March 7, 2008 The freezing point temperatures are correct but I forgot to take in account solubility and ice! Because both Na2CO3 and CaCl2 are very close in that they both produce 3 ions when dissolved and have very close freezing points when comparing in moles, they do not have the same solubility. Solubility of CaCl2 is about 75g/100mL and solubility of Na2CO3 is 30g/100mL therefore I have finally came up with the right solution. I will use CaCl2!!!!!! Link to comment Share on other sites More sharing options...
hermanntrude Posted March 8, 2008 Share Posted March 8, 2008 tell me, what were your original choices? was one of them NaCl? because you can get down to about -20°C with that. I'm not sure that a strong freezing-point depression will help you very much if the reaction isnt endothermic enough. Perhaps you could refer to the CRC handbook and find the deltaH of dissolution for your compounds? Link to comment Share on other sites More sharing options...
swansont Posted March 8, 2008 Share Posted March 8, 2008 tell me, what were your original choices? was one of them NaCl? because you can get down to about -20°C with that. I'm not sure that a strong freezing-point depression will help you very much if the reaction isnt endothermic enough. Perhaps you could refer to the CRC handbook and find the deltaH of dissolution for your compounds? With the mass limit of 20g, NaCl will indeed be better, assuming it all dissolves. The delta-h of the reaction doesn't matter (I had mentioned it earlier but I realize it's immaterial). You have a supply of ice, whose temperature is not specified, that will lower the temperature. Link to comment Share on other sites More sharing options...
hermanntrude Posted March 8, 2008 Share Posted March 8, 2008 there are websites which list the quantities in which you need to mix ice, water and salt to get various different temperatures, for use as ice-baths in temperature sensitive chemical syntheses Link to comment Share on other sites More sharing options...
DivideByZero Posted March 9, 2008 Author Share Posted March 9, 2008 i was correct! it was CaCl2!!! (btw NaCl was not one of the choices) Link to comment Share on other sites More sharing options...
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