Welcome Back Fans!!!

I truly couldn't wait for this moment. The time to write another blog. May I add in 2011? Wow. Speechless. I do not think I can do this. The thrill, the hype, to do something I have not done in a month? Heck, this is going to be one greasy, sloppy, embarrassing blog. All I can say, Preet did it.
Enough about that, I hope everyone else had a fantastic break, and now, it is time to show what exactly when down last class.
Last class we learned material about "Molar Concentration", aka "Molarity of Solutions". It is the concentration measured by the number of moles of solution found in the solution, for every liter. 
The Formula for it is..........M = mol/L
To find the moles, it is........mol = M X L
and for the Volume............L= mol/M

Now of the thousands of you guys who read this blog, I am sure that plenty of you top followers are well aware that I am not a fan of showing examples, but today, why not. Here are 2 fresh from the crew! Actually one, because one is more efficient for both you the reader and me the writer. 


First Example: What is the molarity of 5.30 g of Na2CO3 dissolved in 400 mL of solution?

First you must find the number of moles contained in 5.30 g of Na2CO3
5.30 x 1mol/ 106.0 u =  0.05
Then, since M= mol/L the equation is 0.05/0.4(400 ml in L)= 0.125M

Last Example!!! Only one more to go!!! Wahoooo!!!! 


Ex. 2. How many grams of  Ca(OH)2 are needed to make 100.0 mL of 0.250 M solution?

Find out moles in by using equation mol = ML

= (0.250 M)(0.1 L) = 0.204 mol

Then convert!
0.204 x 74.1/1mol = 1.85 grams of Ca(OH)2

If you're stuck this will help :)
http://www.youtube.com/watch?v=jxSoR8RJDW0&playnext=1&list=PLCDA5A8493EB5AE7E&index=10