Question

17.3.92 g of ferrous ammonium sulphate (Eq. mass = 392) are dissolved in 100 ml of water. 20 ml of this
solution requires 200 ml of potassium permanganate during titration for complete oxidation. The weight
of KMnO4 (Eq. mass = 31.6) present in one litre of the solution is
@ 13.6 g
60.316 g
© 14.76 g
@ 34.78 g​

Answer 1

Explanation:

Molecular weight of Ferrous ammonium sulphate = 392 g/mol

No. of moles of Ferrous ammonium sulphate=

3.92/392=0.01 moles.

Normality of Ferrous ammonium sulphate solution=0.01×1000/100=0.1N

MnO4^- +8H^+ +5e^- =Mn2^+ +4H2OFe2^+=. Fe3+ +e−

From the equations n-factor of (NH4)2 Fe(SO4)2.6H2O=1

and n-factor for KMnO4=5

N1V1=N2V2

0.1×20=N2×18

NE=1/9N

Equivalent weight of KMnO4= Mol. weight/5=158/5=31.6

Weight of KMnO4 present in 1 L of solution =1×31.6/9

=3.476 g

or 34.76 ×10^-1 g

HOPE IT HELPS. (^^)