What weights of Bovine Serum Albumin (molecular weight 66,000) are required to give solutions of 10ml and 100m

Question

What weights of Bovine Serum Albumin (molecular weight 66,000) are required to give solutions of 10ml and 100ml volumes containing 1μmole of BSA

Answer 1

1 umole is a specific quantity.

mole=mass/MW => mass= mole*MW=1*10^-6*66000=66*10^-3 g= 66 mg.

So if you truly want to have 1umole BSA in the final solutions you have to dissolve 66 mg BSA regardless of the volume.

If you meant concentration of BSA umole/volume then you have to define the units that you want; is it umole/Lor umole/mL?

In the first case you need 66 mg in L => 6.6 mg /100 mL and 0.66 mg/10mL. These are extremely small weights and the error is going to be huge. I suggest you make a more concentrated solution and then dilute (use M1V1=M2V2)

In the second case you need 66 mg/mL => 660 mg/10 mL and 6600 mg/100 mL (6.6 g/100 mL).

Answer 2

1 umole is a weight, not a concentration- Its like asking what weight is needed to make 5 pounds. A mole of BSA is 66000 g, thus a millionth of that (umole) is 0.066 g, or 66 mg.
So, 66 mg is 1 umole of BSA.
Now, lets assume you meant to say 1 uMolar, which is a concentration-i.e. moles per liter
66 mg BSA in 1 liter is 1 umole. So is you have 1/10 the volume, you need 1/10 the BSA. So, for a 100ml volume, you need 6.6 mg, and for 10 ml, you need 0.66 mg.