This video is about Colligative Properties - Original
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@theprinterlady8 жыл бұрын
Question - if you have 2.5 moles of LiCl, isn't that 2.5 moles of lithium and 2.5 moles of Chloride = 5 moles of particles? (You used 1 mole each).
@ChemAcademy8 жыл бұрын
You have to take the 2.5 mol of LiCl and use that to find the molality (m). Once you do that you plug that number into the equation for m. The vant hoff factor is 2 since 1 mole LiCl produces 2 moles of dissolved particles in water.
@theprinterlady8 жыл бұрын
The problem states you have 2.5 moles of LiCl. I worked the problem backward, and at 5 moles of particles, it worked out correctly. I'm just not getting how 2.5 moles of LiCl only equals 2 moles of particles... isn't it 2.5 moles of each element? (i.e. 2.5 moles of LiCl = 2.5 moles Li and 2.5 moles of Cl?). Otherwise, 2.5 moles of LiCl = only 1 mole each of the elements? This isn't making sense to me.
@theprinterlady8 жыл бұрын
I should say math is not my strong suit, so I'm easily confused.
@theprinterlady8 жыл бұрын
Here's how I worked it out (won't use proper subscripts). delta T B = Kb x m x i 81.8 - 78.4 = 3.4 change in C Kb - what we are looking for m = moles/kg 2.5 LiCl / 1.75 ethanol i = 2.5 moles of LiCl x 2 = 5 moles particles Kb = 3.4 C / 1.43 x 5 = 0.475 C/m delta T B = Kb x m x i Find for Delta Tb = .476 x 1.43 x 5 = 3.4 If I only use 2 moles particles for i but use 2.5 for the m calc: 2.5 moles LiCl/1.75 L ethanlol = 1.43 m (which agrees with your calc) But then if I only multiply by 2 moles particles I get 1.19 (like you) Here Tb = 1.19 x 1.43 x 2 = 3.4 But I still don't understand where the 2 came from. It works out the same, but I'm not understanding why 2 and not 5 based on the idea that when moles of an ionic compound dissassociate, they form the number of moles of each element.
@theprinterlady8 жыл бұрын
I think I got it = the particle count is always based on one mole (no matter how many moles you have in the problem)?