lab determination of iron in vitamins need help with part B. how do i find the d

Question

lab determination of iron in vitamins

need help with part B.
how do i find the dilution factor of solutions from 1 to 3?

A. Sample tablet extraction, filtration, and dilution how solution I and 2 was made. Obtain a vitamin tablet containing iron and record the label value of the mg Fe per tablet. Pour 25 ml, of 6 MI-ICI into a clean 100- or 150-ml beaker. Add the vitamin tablet, cover with a watch glass, and (in a fume hood) heat gently for 15 minutes. Cool and filter or centrifuge the solution directly into a 100-mL. volumetric flask, washing the beaker and filter several times with distilled water to complete the quantitative transfer. Dilute to the mark with deionized water and mix well. Label this original sample as solution L Dilute the original solution by pipetting 5.00 mL (use 10.00 mL if the label value for iron is less than 15 mg) to 100 mL in a clean volumetric flask. Label this flask solution 2.

Explanation / Answer

Determination of iron in sample

concentration of iron in sample 3 found = 0.626146362 ppm (in 50 ml diluted sample)

dilution factor

concentration of iron in 5 ml sample = 0.626146362 ppm x 50 ml/5 ml

                                                          = 6.26146362 ppm

concentration of iron in 5 ml (original solution) = 6.26146362 ppm x 100 ml/5 ml

                                                                          = 626.146362 ppm

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5 ml original solution diluted to 100 ml : solution 2

5 ml solution 2 diluted to 50 ml : absorbance measured at this stage

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