Asked by ella
0.281g of Fe(NH4)2(SO4)2*6H2O is dissolved in a total volume of 500.0 mL of water, calculate the resulting Fe concentration in ppm.
molar mass of Fe(NH4)2(SO4)2*6H2O = 392.14088g/mol
molar mass of Fe = 55.845g/mol
0.281g * (1mol/392.14088g) = 7.165X10^-4 mol
7.165X10^-4 mol * (1mol Fe/ 1mol Fe(NH4)2(SO4)2*6H2O) = 7.165X10^-4 mol Fe
7.165X10^-4 mol (55.845g Fe/ 1 mol Fe) = 0.040017g Fe
0.040017g Fe / 500.0mL = 8.003X10^-5g/mL
(8.003X10^-5g/mL)(1000mL/L) = 0.08003g/L
(0.08003g/L)(1000mg/g) = 80.0mg/L
80.0mg/L = 80.0 ppm since 1mg/L = 1ppm
molar mass of Fe(NH4)2(SO4)2*6H2O = 392.14088g/mol
molar mass of Fe = 55.845g/mol
0.281g * (1mol/392.14088g) = 7.165X10^-4 mol
7.165X10^-4 mol * (1mol Fe/ 1mol Fe(NH4)2(SO4)2*6H2O) = 7.165X10^-4 mol Fe
7.165X10^-4 mol (55.845g Fe/ 1 mol Fe) = 0.040017g Fe
0.040017g Fe / 500.0mL = 8.003X10^-5g/mL
(8.003X10^-5g/mL)(1000mL/L) = 0.08003g/L
(0.08003g/L)(1000mg/g) = 80.0mg/L
80.0mg/L = 80.0 ppm since 1mg/L = 1ppm
Answers
Answered by
DrBob222
That looks good to me. Good work.
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