- #1
higherme
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here is the question:
The entire contents of one of the Erlenmeyer flasks used in the oxalate-permanganate titration in Experiment 8 was poured into a 250 mL volumetric flask. The volume was made up to 250 mL. A 20 mL portion of the solution was placed in a 100 mL volumetric flask and 20 mL 0.01 M thiocyanate, and 60 mL 0.5 M nitric acid was added. The absorbance of the solution was measured and the intersection plotted on the calibration graph. The concentration of Fe(III) in the solution was found to be 2.410E-4 M. Calculate the mass of iron in the Erlenmeyer flask from Experiment 8. Enter your answer in scientific notation with three significant figures.
My attempt:
calculate the amount of moles in that 100mL flask. so,
since we know the concentration of FeIII is 2.410E-4 M
I multiplied that by the volume of the flask which is 100 mL = 0.1 L
I got 2.41E-5 moles of FeIII
Then i took that mole of FeIII and multiplied it by molar mass of FeIII which is 55.845 g/mol ( unless there is another mass because i know there are Fe II and Fe III)
now i got 1.35 E-3 g of FeIII which is in the the original erlenmeyer flask??
well, my answer was wrong according to the computer.
The entire contents of one of the Erlenmeyer flasks used in the oxalate-permanganate titration in Experiment 8 was poured into a 250 mL volumetric flask. The volume was made up to 250 mL. A 20 mL portion of the solution was placed in a 100 mL volumetric flask and 20 mL 0.01 M thiocyanate, and 60 mL 0.5 M nitric acid was added. The absorbance of the solution was measured and the intersection plotted on the calibration graph. The concentration of Fe(III) in the solution was found to be 2.410E-4 M. Calculate the mass of iron in the Erlenmeyer flask from Experiment 8. Enter your answer in scientific notation with three significant figures.
My attempt:
calculate the amount of moles in that 100mL flask. so,
since we know the concentration of FeIII is 2.410E-4 M
I multiplied that by the volume of the flask which is 100 mL = 0.1 L
I got 2.41E-5 moles of FeIII
Then i took that mole of FeIII and multiplied it by molar mass of FeIII which is 55.845 g/mol ( unless there is another mass because i know there are Fe II and Fe III)
now i got 1.35 E-3 g of FeIII which is in the the original erlenmeyer flask??
well, my answer was wrong according to the computer.