1. The molecular weight of ammonium sulfate is 132.14 g mol -1 and you have a so

Question

1. The molecular weight of ammonium sulfate is 132.14 g mol-1 and you have a solution of total volume 9 mL. What weight of ammonium sulfate should you add to reach a final concentration of 1.5 M (in grams to 2 s.f.)?

2.What is the pH of the buffer you will use in the anion-exchange chromatography purification step in your purification of pyruvate kinase?

3.You have a solution with a protein concentration of 12 mg mL-1. The total volume of this solution is 18 mL. What is the total amount of protein (in mg to 2 s.f.)?

Explanation / Answer

1. mol. wt. = 132.14g/mol

vol. = 9/1000 L

Concentration(in M) = no. of moles/ vol. in L

no. of moles = 1.5 x 9/1000 = 0.0135

wt. of substance = no. of moles x mol. wt. = 0.0135 x 132.14 = 1.7839 g

2. In anion exchange chromatography, the protein is separated based on the negative charge they have. any protein will have a negative charge at a pH > pI. So, in this case, the pH of the loading buffer should be such that it is greater than the pI of Pyruvate Kinase.

3. Concentration = 12 mg/mL

Vol. = 18mL

Concentration (in mg/mL) = wt./ vol.

wt. = 12 x 18 = 216 mg

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