Question

Question #16097

Answer 1

`"Molar mass of "KMnO_4`
`=39+55+4*16=158g/"mol"`

`0.02(M) KMnO_4`
`=0.02"mol"/Lxx158g/"mol"=3.16g/L" "KMnO_4`

So to make 1L 0.02M solution(secondary standard) approximately 3.16g of `KMnO_4` is taken in a 1L volumetric flask and then dissolved in water to make volume 1L.