Had a 100 mL flask containing 59.96 milligrams of iron the results of lab 6. ___

Question

Had a 100 mL flask containing 59.96 milligrams of iron the results of lab 6.

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using 1 mL of the iron unknown sample from lab 6.

4 mL of hydroxylamine HCL

40-mL of 0.1 M of sodium acetate solution

40-mL of the 1, 10 phenanthroline solutions

DI water (to dilute these solutions to a desired volume of 1 Liter or 1000 mL), hence about 915 mL of DI water.   

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Ran a spectometry test and discovered in the test tube there was 3.675 mg/L.

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Determing the iron content

Explanation / Answer

Ans. Given-

I. 1.0 mL unknown solution is diluted to 1000.0 mL (= 1.0 L). Let’s call it solution 1.

II. 1.0 mL aliquot of solution 1 gives [Fe] = 3.675 mg/L

Conclusion: Note that all [Fe] in solution 1 is solely due to 1.0 mL original unknown solution in it. So, total amount of Fe in 1.0 L of solution 1 is equal to total Fe in 1.0 mL original unknown solution.

Now,

Total amount of Fe in solution 1 = [Fe] of solution 1 x Vol. of solution 1

= (3.675 mg/ L) x 1.0 L

= 3.675 mg

Therefore, the total amount of Fe in solution 1 = 3.675 mg

Also,

Amount of Fe in 1.0 mL original unknown = total amount of Fe in solution 1

Hence, Amount of Fe in 1.0 mL original unknown = 3.675 mg

# Concertation of Fe in unknown solution =

Amount of Fe in original unknown / vol. of original unknown

= 3.675 mg/ 1.0 mL

= 3.675 mg/ mL

= 367.5 mg/ 100 mL

= 3675 mg/ L                                               ; [1 mL = 0.001 L]

= 3.675 g/ L                                                 ; [1 g = 1000 mg]

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