Question

6.95gm of FeSO4.XH2O crystals was accurately weighed and dissolved in dil. H2SO4. This solution was diluted upto 250 cc. 20 cc of this solution needed 16 cc of N/8 KMNO, solution for complete reaction. Calculate the value of x.

Answer 1

Answer:

6.95gm of FeSO4.XH2O crystals was accurately weighed and dissolved in dil. H2SO4. This solution was diluted upto 250 cc. 20 cc of this solution needed 16 cc of N/8 KMNO, solution for complete reaction.

Answer 2

The value of X in FeSO4.XH2O is 7.

• Given, Mass of FeSO4.XH2O taken = 6.95 gm
• Volume of FeSO4.XH2O solution = 250 ml
• Also, in titration,    KMNO4 normality = N/8
• Volume of KMNO4 required for complete titration = 16ml
• Volume of FeSO4.XH2O solution used in titration = 20 ml
• Now, the reaction is.....
• MnO4- + 8 H+ + 5 Fe2+  ⇒  Mn2+ + 4 H20 + 5 Fe3+
• Here KMnO4 is reduced to Mn+2, number of electrons transferred = 5
• Also, Fe2+ is oxidized to Fe3+, number of electrons transferred = 1
• The ratio between Fe2+ and MnO4- = 5:1
• Therefore, to find Normality of FeSO4.XH2O used we have                
• N1V1 = N2V2
• 1*N1*20 = 5*1/8 * 16 =  N1 = 0.1N
• Now we know to calculate mass of FeSO4.XH2O in 40N FeSO4.XH2O Solution we have,
• Normality = Weight / Gram.Equivalent * 1000/Volume in ml
• 0.1 = Weight / 152 * 1000 / 250
• Weight = 3.8 gm
• Therfore, moles FeSO4 in the solution = 3.8/151.908 =0.025
• Now if the solution has 3.8 gm of FeSO4 in 6.95 gm FeSO4.XH2O, then the remaining mass will be of H2O molecules.
• Weight of H2O in the solution = 6.95 - 3.8 = 3.15 gm
• Moles Of water = 3.15/18.02 = 0.175
• Now molar ratio of H2O : FeSO4 = 0.175 / 0.025 = 7
• Therefore, value of X = 7