Question

Weigh accurately about 0.0702 g of pure ferrous ammonium sulphate hexahydrate, dissolve it in distilled water & transfer the solution to a 1liter volumetric flask. Add 2.5mL of conc. sulfuric acid & dilute the solution to the mark with distilled water. Calculate the concentration of solution in mg of Fe / mL.

Answer 1

Molar mass of Ferrous ammonium sulphate hexahydrate(FAS.6H2O) is 392.14

Atomic mass of Fe 55.85 g / mol

i.e. 392.14 g of FAS.6H2O   55.85 g Fe

Hence, 0.0702 g of FAS.6H2O say 'M' g Fe.

M = (55.85 x 0.0702) / (392.14)

M = 3.921 / 392.14

M = 9.998 x 10^-3 g

It means, M = 9.998 x 10^-3 g Fe in 1000 mL solution.......(As 1 L of solution prepared)

i.e. M = 9.998 x 10^-3 g / 1000 mL

Hence, M = [(9.998 x 10^-3) / 1000 ] g / mL.....(divided by 1000 to make it per 1mL)

M = 9.998 x 10^-6g/mL

M = 9.998 x 10^-6 x 10³mg/mL of Fe .............( 1g = 10³ mg)

M = 9.998 x 10^-3 mg/mL.of Fe

Hence concentration of solution in mg of Fe / ml is 9.998 x 10^-3 mg/mL