100mL of a solution that is simultaneously 15.6 mg/mL malonic acid, 3.38 mg/mL MnSO4 • H2O, and 0.03% starch

Question

Calculate the volumes necessary to make this solution in g and mL

Answer 1

To calculate the volumes necessary to make this solution, we need to know the mass of each solute that was added to the solution.

Assuming that the 100 mL of solution contains 15.6 mg/mL malonic acid, we can calculate the mass of malonic acid in the solution as follows:

Mass of malonic acid = 15.6 mg/mL x 100 mL = 1560 mg

Assuming that the 100 mL of solution contains 3.38 mg/mL MnSO4 • H2O, we can calculate the mass of MnSO4 • H2O in the solution as follows:

Mass of MnSO4 • H2O = 3.38 mg/mL x 100 mL = 338 mg

Assuming that the 0.03% starch is w/v (weight/volume), we can calculate the mass of starch in the solution as follows:

Mass of starch = 0.03/100 x 100 mL x 1 g/mL = 0.03 g

Therefore, the total mass of solutes in the solution is:

1560 mg + 338 mg + 0.03 g = 1898.03 mg

To convert the mass of the solution to volume, we need to know the density of the solution. Without this information, we cannot convert the mass to volume.