How many nanomoles (nmol) of bovine serum albumin (BSA) would be present in 7.2 mL of...
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?
Homework Answers
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
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