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Finding Unknown Angles

K2CrO4 oxidiz...

Question

$K_{2} C r O_{4}$ oxidizes $K I$ in the presence of $H C l$ to $I_{2}$ . The equivalent weight of the $K_{2} C r O_{4}$ is:

\frac{M w}{2}

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M w \times \frac{2}{3}

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\frac{M w}{3}

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\frac{M w}{6}

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Solution

The correct option is C $\frac{M w}{3}$
$3 e^{-} + C r O_{4}^{2 -} x = 6 \to [C r (O H)_{4}]^{⊝} x = 3 (x = 6 - 3 = 3)$
$∴ E w = \frac{M w}{3}$

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2 C u S O_{4} + 4 K I \to C u_{2} I_{2} + 2 K_{2} S O_{4} + I_{2}

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{\to v}_{m}

θ

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