I cant seem to get the answer Finally, a vial containing a sample of finely powd

Question

I cant seem to get the answer

Finally, a vial containing a sample of finely powdered iron ore is found to weigh 9.8117g After most of the ore sample was poured into a beaker, the nearly empty vial weighed 9.6075 g. The ore is digested in HCI and diluted to 100.0 mL. A 25.00 mL aliquot of this sample solution is diluted to 1000.0 mL using volumetric glassware. Then, a 5.000 mL aliquot of this diluted sample solution is transferred to a 100.0 mL volumetric flask, buffered to pH 3.5, reduced with H2Q, complexed with phen, and diluted to the mark. The absorbance of this unknown test solution is read, and a calibration curve is used to determine its concentration to be 0.87472 mg/L. what is the wt% of iron in the ore? Iron ore: 34.269% Fe]

Explanation / Answer

We will solve the problem step by step

Initial weight of vial + ore = 9.8117g

Weight of vial = 9.6075

Weight of ore = 9.8117 - 9.6075 = 0.2042 grams

Volume to which it is diluted initially = 100mL

so ore / 10mL = 0.2042g /100mL

The aliquot taken = 25mL

So ore in 25mL = 0.2042 X 50 / 100= 0.05105 grams

diluted to 1000mL so amount of ore per 1000mL = 0.05105grams

volume of aliquot from 1000mL = 5mL

Amount of ore in this 5mL = 0.05105 X 5 /1000 = 0.00025525grams

this is further diluted to 100mL

so amount of ore in the futher diluted 100mL = 0.00025525grams = 0.25525mg / 100mL

The concentration of iron found to be = 0.87472mg / L = 0.087472mg / 10mL

Percentage of iron = 0.087472 X 100 / 0.25525 = 32.269%

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