1) You have 100 ml of a solution of ATP. You take 50 µl, and dilute it to a fina

Question

1) You have 100 ml of a solution of ATP. You take 50 µl, and dilute it to a final volume of 2000 µl. The absorbance of the diluted solution at 259 nm is 0.520.

a) What is the ATP concentration in the original, undiluted solution? (5 points) The molar extinction coefficient for ATP at 259 nm is 15,600 l mol-1cm-1.

b. How many mg of ATP are there in total in the original, undiluted solution? The formula weight of ATP is 551.14 g/mole. (5 points)

The last time this question was posted the answer was wrong, so please don't copy and paste the wrong answer. Thanks.

Explanation / Answer

1) A = e x C x L

where A = absorbance = 0.52 , e = 15600 , L = 1cm ( in general L = 1cm)

hence we find C = concentration

0.52 = 15600 x C x 1

C = 0.000033

volume of solution = 2000 ul = 2000 x 10^-6 L = 2 x 10^-3 L

Moles of ATP = M x V = 3.33 x 10^-5 x 2 x 10^-3 = 6.66 x 10^-8

This is present in undiluted 50 ul solution , volume = 50 ul = 50 x 10^-6 L = 5 x 10^ -5 L

Concentration of undiluted solution = moles / Volume = 6.66 x 10^-8 / 5x10^-5 = 0.00133 M

Since this solution is part of orginial solution

Now concentration of solution with 100 ml ( i.e initial) = 0.00133 M

b) Moles of ATP in original solution = M x V = 0.00133 x 0.1 = 0.000133

Mass of ATP = moles x molar mass of ATP = 0.000133 x 551.14 g/mol = 0.0735 g = 73.5 mg

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