Chromel is an alloy composed of nickel, iron, and chromium. A 0.6472 g sample wa

Question

Chromel is an alloy composed of nickel, iron, and chromium. A 0.6472 g sample was dissolved and diluted to 250.0 mL When a 50 00-mL aliquot of 0 05182 MEDTA mixed with an equal volume of the diluted sample, all three ions were chelated and a 5.11-ml. back-titration with 0.06241 M copper (II) was required. The chromium in a second 50.0-mLaliquot was masked through the addition of hexamethylenetetramine. titration of the Fe and Ni required 36.28 mL of 0.05182 MEDTA Iron and chromium were masked with pyrophosphate in a third 50.0-mL aliquot, and the nickel was titrated with 25.91. mL of the EDTA solution Calculate the percentages of nickel, chromium, and iron in the alloy.

Explanation / Answer

Alloy = 0.6472 g

moles of EDTA added to 50 ml aliquot = 0.05182 M x 50 ml = 2.591 mmol

moles of Cu(II) used for back titration of EDTA used = 0.06241 M x 5.11 ml = 0.32 mmol

moles of EDTA used by Ni+Fe+Cr = 2.59 -0. 32 = 2.27 mmol

Total moles of Ni+Fe+Cr in 250 ml sample solution = 2.27 mmol x 250 ml/50 ml = 11.35 mmol

EDTA used for 50 ml aliquot (Cr masked) = 0.05182 M x 36.28 ml = 1.88 mmol of Fe+Ni

Fe + Ni in 250 ml sample solution = 1.88 mmol x 250 ml/50 ml = 9.4 mmol

moles of Cr in sample solution = 11.35 - 9.4 = 1.95 mmol

Percentage of Cr in alloy = 1.95 mmol x 51.996 g/mol x 100/1000 x 0.6472 = 15.66%

EDTA used for 50 ml aliquot (Fe+Cr masked) = 0.05182 M x 25.91 ml = 1.34 mmol Ni

Ni in 250 ml sample solution = 1.34 mmol x 250 ml/50 ml = 6.7 mmol

percentage of Ni in alloy = 6.7 mmol x 58.69 g/mol x 100/1000 x 0.6472 g = 60.76%

Percentage of Fe in alloy = 100 - 15.66 - 60.76 = 23.58%

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