1.

Equivalent weight of K_(2)Cr_(2)O_(4) in acidic medium is

Answer»

`("MOL. wt")/(3)`
`("Mol. wt.")/(6)`
`("Mol. wt.")/(2)`
`"Mol.wt."`

Solution :`14H^(+)+Cr_(2)O_(7)^(2-)+6E^(-)rarr2Cr^(3+)+7H_(2)O`
`"Eq. wt"=("Mol. wt.")/("No. of electrons GAINED")=("Mol. wt.")/(6)`


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