3.47 g of the hydrated "double salt", ammonium iron(II) sulfate hexahydrate, FeSO4(NH4)2SO4•6H2O was

1. Molarity of salt :

M = mole of salt / Volume of solution (in L)

M = n/V (in L)

where n = mass of salt /molar mass of salt

mass of salt = 3.47 g, Molar mass of salt = 392.13 g/mol,

Volume = 200 ml x 1L/1000mL =0.2 L

there fore 

M_salt  = (3.47 g / 392.13 g/mol)   x  1/ 0.2 L

          = 0.044 M

In redox reaction Fe⁺² is converted into Fe⁺³  Change in oxidation number of Fe is 1 (1 unit increased)

Hence valence factor is 1

there fore 

Normality of salt (N_salt) = M_salt  x Valence factor

                                       = 0.044  x 1

                                       = 0.044 N