How many nanomoles (nmol) of bovine serum albumin (BSA) would be present in 7.2 mL of a 1.4 mg/mL solution of BSA?
As given in the question:
Amount of BSA present in 1 ml of solution = 1.4 mg
Total volyme of solution = 7.2 ml
therefore Amount of BSA present in 7.2 ml of Solution= 1.4 x 7.2 = 10.08 mg
Molecular mass of 1 mole of BSA = 66430 g
66430 g of BSA represents 1 mole
1g represents 1/66430 moles
10.08 g represent 1/66430 x 10.08 moles
= 0.0000012042 moles
=1204.2 n mol
How many nanomoles (nmol) of bovine serum albumin (BSA) would be present in 7.2 mL of...
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