When CrI_3 oxidises to Cr_2O_7^{2-} and IO_4^ | vedclass

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(B) The oxidation states change as follows: $\mathop{CrI_3}\limits^{(+3)(3 	imes -1)} 	o \mathop{Cr_2O_7^{2-}}\limits^{(2 	imes +6)} + \mathop{3IO_

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$\frac{M}{27}$

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(B) The oxidation states change as follows: $\mathop{CrI_3}\limits^{(+3)(3 	imes -1)} 	o \mathop{Cr_2O_7^{2-}}\limits^{(2 	imes +6)} + \mathop{3IO_4^-}\limits^{(3 	imes +7)}$.For $Cr$: The change in oxidation state is $(+6 - (+3)) = +3$ per $Cr$ atom.For $I$: The change in oxidation state is $(+7 - (-1)) = +8$ per $I$ atom. Since there are $3$ iodine atoms,the total change is $3 	imes 8 = 24$.Total $n$-factor $= 3 + 24 = 27$.Equivalent mass $E = \frac{M}{n	ext{-factor}} = \frac{M}{27}$.

When $CrI_3$ oxidises to $Cr_2O_7^{2-}$ and $IO_4^-$,the equivalent mass of $CrI_3$ will be :-

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