March 21st, 2011

We started off today's class reviewing Friday's blog and then going right to the reading sheet corrections, which Mr.H went through in under 2 minutes (classic Mr.H). The answers for pages 23-24 are as follows:
1.a,b,c,d
2.a
3.b
4.b
5.a
6.c
7.Didn't get this one >.<
8. c
9.c
10.a
11.d
12.a
13.c
14a. 0.645 molal
14b.0.225kg
14c.34.45g

After going over these answers, we continued on to Page 5 which is really really really awesome for learning how to set up equations to find Molarity, Molality, Mass Percentage and Volume Percentage. Every equations is given to you in its general form and I highly recommend using this sheet to study for the test coming on Friday. Anyways, we did problems 1,2,3 and 6 on pages 5 and 6. If you got problem 1, you will get the rest of them for sure. But, if you didn't, this is how to go about doing problem 1:
First, get all of your general information out of the way. What I mean by this is calculate moles of CaCl2, kg and L of solvent (water), and mass of solution. The reason that we do this first is because we are looking for Molarity, Molality, and mass percentage in the given solution. If you get all of this general information out of the way, all you will have to do is plug in this information into their respective formulas. Next, we calculated Molarity. By using the chart at the top of the page, we know that this is moles of solute/ L of solution, which, if u followed the steps prior, you will be able to plug in no problem. The answer to the molarity is .284M. Then we calculated Molality. We see that this has the same numerator as the molarity so we can use the same moles of solute that we used for molarity. We also see that the denominator of molality is kilograms of solvent. Once again, we can just plug in the general information that we got at the beginning of the problem. We should get the exact same answer as we did for molarity, but with a molal after it of course (the answer is .284molal). Finally, we calculate the mass %. This is a very simple process in that, even if you didn't get all of the general information we talked about, the answer is pretty much given to you in the question. Just place mass of CaCl2 (given) over mass of CaCl2 + the mass of solvent (100mL or 100g). You should have gotten 3.15/103.15 or 3.05%. If you have gotten this far you deserve a puppy.



(this is Boo. he is happy for you.)
Why the puppy? Well, its because you are about to do question 6 (actually not that difficult) and the Webassign (actually that difficult). When looking at number 6, it is important to assume 1L of solution throughout the question. This will give us a nice, round relative number to work with. Also, it gives you the number of moles of sulfuric acid (H2SO4) which is important. This will give us the mass of solute after we multiply 3.75 mols by the molar mass. You should have gotten 367.75 g of H2SO4. Now, because we assumed 1L of solution, we can just multiply the given density by 1000 to get the mass of solution. Now we can solve the first part of question 6; the mass percent. We will set up just like we did in question , but this time DO NOT ADD THE SOLUTE TO YOUR DENOMINATOR. We don't add it in this case because we already have the mass of solution (not solvent) which has already taken into account the solute and solvent. The answer to mass % is 29.9%. Finally, we calculated molality by placing the given amount of moles over the kilograms of solvent. We find the kg of solvent by multiplying 1230 by 70.1%. We use this percentage because it is what is left over after we calculate the mass of solute in the mass%. So, you should get 4.35 molal.

Congratulations you have finished the "math-intensive part of the Solution Unit. Sadly, you have to do the webassign now which used the same equations that we used on pgs 5 and 6, but with more complicated wording >.<. Good Luck!





(but hey. atleast your not doing this?)