Asked by BabyMikee
Describe the preparation of 100 mL 0.50 M solution of Fe(NO3)3 from a stock solutiion that is 60%(w/w)Fe(NO3)3 solution with a specific gravity of 1.110 g/ml. Calculate the molar NO3 concentration in the resulting solution. Molar Mass of Fe(NO3)3: 241.86
Answers
Answered by
DrBob222
First I would calculate the molarity of the iron(III) nitrate solution.
1.110 g/mL x 1000 mL x 0.60 = g/L
Then g/molar mass = mols and this is the M of the soln.
Then dilute that stock soln using the dilution formula.
c1v1 = c2v2 where
c = concn
v = volume.
1.110 g/mL x 1000 mL x 0.60 = g/L
Then g/molar mass = mols and this is the M of the soln.
Then dilute that stock soln using the dilution formula.
c1v1 = c2v2 where
c = concn
v = volume.
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