1)
a)
Molar mass of Fe(CH3COO)3,
MM = 1*MM(Fe) + 6*MM(C) + 9*MM(H) + 6*MM(O)
= 1*55.85 + 6*12.01 + 9*1.008 + 6*16.0
= 232.982 g/mol
mass(Fe(CH3COO)3)= 16.6 g
number of mol of Fe(CH3COO)3,
n = mass of Fe(CH3COO)3/molar mass of Fe(CH3COO)3
=(16.6 g)/(232.982 g/mol)
= 7.125*10^-2 mol
volume , V = 125 mL
= 0.125 L
Molarity,
M = number of mol / volume in L
= 7.125*10^-2/0.125
= 0.570 M
Answer: 0.570 M
b)
[Fe3+] = [Fe(CH3COO)3] = 0.570 M
Answer: 0.570 M
c)
[CH3COO-] = 3*[Fe(CH3COO)3] = 3*0.570 = 1.71 M
Answer: 1.71 M
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