chem biochem bio
posted by mel .
You have the following stock solutions and solid chemicals in your lab:
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???
0.05M Tris, pH 8.0
0.25 mg/ml BSA
If i understand the NaCL can be determined by c1v1=c2v2 but the BSA MgCL2 and Tris I don't understand, plus is the total suppossed to add to be 250 ml?
NaCl, BSA, and Tris are dilutions that use c1v1 = c2v2. There is no difference in the procedure for Tris and NaCl.
For BSA solution, you have a concn for that which isn't M but mg/mL. Same process.
The MgCl2 is not a dilution problem, per se, but it is diluted in the end.
For the MgCl2 find grams (or milligrams) MgCl2 required to make 250 mL of 25 mM MgCl2, from mols = M x L. You know M and L, solve for mols. Then mols = grams/molar mass. You know molar mass and mols, solve for grams. Then since the stuff is only 95% pure, divide that number by 0.95.
I'm not sure what total you are talking about for 250 mL. The FINAL solution must be 250 mL. The way you do that is to put grams/mL of the stock solution into a 250 mL volumetric flask and make to the mark with DI or distilled H2O. That way the final solution will ber 250 mL.