1.

In acidic medium, the equivalent weight of K_(2)Cr_(2)O_(7)(molecular weight = M) is

Answer»

M
M/2
M/3
M/6

Solution :Equivlent WEIGHT `=("Molecular weight")/("CHANGE in oxidation state/valency/no. of electron involved")`
Reaction involved is
`Cr_(2)O_(7)^(2-) + 14H^(+) + 6e^(-) to 2Cr^(3+) + 7H_(2)O`
It is quite clear that in this reaction, `Cr(+6)` is reduced to `Cr(+3)`.
`therefore` Equivalent weight of `K_(2)Cr_(2)O_(7) = ("MOL. wt")/6 = M/6`


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