An inhibitor of molecular weight 450 is given to you as an accurately weighed 12
ID: 932778 • Letter: A
Question
An inhibitor of molecular weight 450 is given to you as an accurately weighed 12 mg aliquot of dry powder need to prepare a stock solution at 1.5 mM final concentration. You do not have access to a balance. Describe what you do. Write your answers in concise and clear steps: 1. Do this 2. Do that etc. You are given a solution containing a pure protein and you are told the extinction coefficient at 280 nm of the protein is 29,000 M^-1 cm^-1. The solution has an absorbance at 280 nm of 1.7. What is the concentration?Explanation / Answer
Given concentration of stock solution , M = 1.5 mM = 1.5x10-3 M
weight of the inhibitor , w = 12 mg = 12x10-3 g
Molecular weight = 450
So number of moles of inhibitor , n = mass/ molecular weigth
= (12x10-3 )/ 450
= 2.67x10-5 mol
We know that Molarity , M = number of moles / volume of solution in L
So volume of solution in L , V = number of moles / molarity
= (2.67x10-5 mol) / (1.5x10-3 )
= 0.0178 L
= 0.0178x1000 mL
= 17.8 mL
Therefore after weighing & transfer exactly 12 mg of the inhibitor into a flask & inorder to prepare 1.5mM of solution dilute the solution by adding 17.8 mL of water.
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WE know that absorbance , A = abc
Where
a = extinction coefficient = 29000M-1cm-1
b = path length = length of the cuvette = 1cm
c = concentration = ?
A = absorbance = 1.7
Plug the values we get
c = A/(ab)
= 1.7 / (29000x1)
= 5.86x10-5 M
Therefore the concentration of the solution is 5.86x10-5 M