The amount of KMnO4 required to prepare 100 ml of 0.1N solution in alkaline medium is:
(a) 1.58 gm (b) 3.16 gm (c) 0.52 gm (d) 0.31 gm
Answer: (a) In alkaline medium KMnO4 act as oxidant as follows:
2KMnO4 + 2KOH = 2K2MnO4 + H2O + (O)
Hence its eq. wt. = m. wt. = 158
Now, Normality = (mass/Eq. mass) x 1/V(L)
mass = 0.1 x 158 x (100/1000)g = 1.58 g