1.

In the neutralization of `Na_(2)S_(2)O_(3)` using `K_(2)Cr_(2)O_(7)` by idometry, the equivalent weight of `K_(2)Cr_(2)O_(7)` isA. (a)`"molecular weight"//2`B. (b)`"molecular weight"//6`C. (c )`"molecular weight"//3`D. (d)same as molecular weight

Answer» Correct Answer - B
In this reaction
`Cr_(2)O_(3)^(2-)` is changed into `Cr_(3)^(3+)` ion.
So, equivalent weight of
`K_(2)Cr_(2)O_(7)=("molecular wt".)/("decrease of oxidation number"xxn)`
`("where n= No. of atoms of" Cr)`
`=("molecular wt".)/(3xx2)=("molecular wt.")/6`


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