Today’s class was rather blue. That is, the solutions were. But, they had different shades, and of course, this was no coincidence. We were told these were solutions of copper (II) chloride, or cupric chloride.
Our task was to figure out which of the solutions corresponded to 0.1 M. No problem, we said! We decided to quickstyle this lab.
The items we worked with included the following:
Copper (II) chloride
Scoopula
Weight
Plastic container
Beaker
Erlenmeyer flask
Graduated cylinder
Here is the procedure:
1. Obtain your needed materials listed above.
2. Weight out 1.345g of CuCl2.
3. Measure 100mL of water in a graduated cylinder.
4. Mix these two substances in an Erlenmeyer flask (it’s fun to spin).
5. Pour into test tube and compare with test tube species.
How we obtained 1.345g?
Simple. To receive a molarity of 1.0, we know that there are 1.0 moles of CuCl2 in 1L. To receive our desired molarity, 0.1, we have 0.1 in 1L. Since one liter is a bit too much, we simply keep our ratio the same, of moles and liters, but decrease the numbers by 10. Now we have 0.01 moles in 100 mL. How many grams is this? Simply molar mass of CuCl2 will tell us.
0.01mol CuCl2 (134.5g / 1 mol) = 1.345g Tada :)
Posted by Andrew.
No comments:
Post a Comment