The equivalent weight of $K_{2} {C r}_{2} O_{7}$ in acid medium is equal to:

molecular weight

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1 / 2

molecular weight

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1 / 6

molecular weight

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1 / 5

molecular weight

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Solution

The correct option is C $1 / 6$ molecular weight
In acidic medium, there is an exchange of 6 electrons hence the equivalent weight of $K_{2} C r_{2} O_{7}$ is $\frac{1}{6}$ of its molecular weight.

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Similar questions

K_{2} {C r}_{2} O_{7} - -------- - + 7 H_{2} S O_{4} + 6 F e S O_{4} ⟶ K_{2} S O_{4} + {C r}_{2} {(S O_{4})}_{3} + 3 {F e}_{2} {(S O_{4})}_{3} + 7 H_{2} O

M

{C r}_{2} O_{7}^{2 -}

{C r}^{3 +}

{C r}_{2} O_{7}^{2 -} = x \times

x

I_{2} + 2 S_{2} O_{3}^{2 -} \to 2 I^{-} + S_{4} O_{6}^{2 -}

K_{2} C r_{2} O_{7}

F e S O_{4}

(N H_{4}) 2 S O_{4} .6 H_{2} O

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