biochem
posted by Shea .
You have the following stock solutions and solid chemicals in your lab:
1.5M NaCL
0.2M Tris buffer, pH 8.0
10 mg/ml Bovine serum albumin (BSA)
MgCl2 (solid, anhydrous) MW = 95.211 g/mole (95% pure)
How would you make 250 ml of the following solution:
0.2M NaCl
0.05M Tris, pH 8.0
0.25 mg/ml BSA
25mM MgCl2

There are just dilutions. You can use the dilution formula for most if not all.
c1v1 = c2v2
c = concn
V = volume
1.5M x v1 = 0.2M x 250 mL
v1 = 0.2 x 250/1.5 = 33.33.
So you place 33.33 mL of the 1.5M stock solution in a 250 mL volumetric flask, add water to the mark on the flask and mix thoroughly.
Someone will be happy to check your answers for the others if you care to post them. 
I am confused on the dillution of the MgCl2 do I have to divide by 95 because it is only 95% pure?

Yes, if you divide the right number by the right number. :)
Calculate the grams MgCl2 you need, then divide that number by 0.95 to find the final grams to use.