Chromel is an alloy composed of nickel, iron, and chromium. A 0.6472-g sample is
ID: 529856 • Letter: C
Question
Chromel is an alloy composed of nickel, iron, and chromium. A 0.6472-g sample is dissolved and diluted to 250.00 mL. When a 50.00-mL aliquot of 0.05182 M EDTA was 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) solution was required. The chromium in a second 50.00-mL aliquot was masked through the addition of hexamethylenetetramine and the titration of the Fe and Ni in the sample required 36.28 mL of 0.05182 M EDTA. Iron and chromium were masked with pyrophosphate in a third 50.00-mL aliquot an the nickel was titrated with 25.91 mL of the EDTA solution. Compute the mass percentages of Ni, Fc and Cr in the alloy.Explanation / Answer
5.11 ml-------------Cu2+ 0.06241 M
50 ml Ni, Fe, Cr +50 ml of 0.05182 M EDTA
36.28 ml ----- EDTA 0.05182 M
50 ml Ni, Fe, Cr
MVEDTA = MVNi+ MVF
0.05182 x 36.28= M50Ni + M50Fe
MFe= 0.0108 M
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25.91 ml ----EDTA 0.05182
M 50 ml
MVEDTA = MVNi
0.05182 x 25.91 = M x 50
M Ni=0.0268 M
M VCu = M V unreacted EDTA
0.06241 x 5.11 = 0.05182 x V
V unreacted EDTA = 6.15ml
V of EDTA reacted with Ni, Fe and Cr = 50 – 6.15 = 43.85 ml
(MV)Ni + (MV)Fe + (MV)Cr= (MV)EDTA
(M50)Ni + (M50) Fe + (M50) Cr = (0.05182 x 43.85) EDTA
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MFe= 0.0108 M
MNi=0.0268 M
MCr= 7.8 x 10-3 M
After calculating the molarities, determine the weight of each metal in the sample.
. Wt of Ni = (0.0268×250×58.693)/1000 = 0.393 g
Wt of Fe = (0.0108×250×55.845)/1000 = 0.151 g
Wt of Cr = (7.8×10-3×250×51.996)/1000 = 0.101 g
the percentage of each metal in the sample.
% Ni = (0.393/0.6472)x 100 = 60.72 %
% Fe = (0.151/0.6472)x 100 = 23.33 %
% Cr = (0.101/0.6472)x 100 = 15.61%
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