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Q.

It takes 0.15 $m o l$ of $C l O^{-}$ to oxidize 12.6g of chromium oxide of specific formula of $C r_{2} O_{x}$ to $C r_{2} {O_{7}}^{- 2}$ , $C l O^{-}$ become $C l^{-}$ . The formula of the oxide is (Atomic weight of $C r = 52, O = 16$ )

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a

$C r O_{2}$

b

$C r O_{4}$

c

$C r O_{3}$

d

$C r O$

answer is B.

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Detailed Solution

$C l o^{-} + C r_{2} o_{x} \to C l^{-} + C r_{2} {O_{7}}^{- 2}$
   $2 (6 - x)$
Number of equivalent of $C l o^{-} =$ Number of equivalent of    $C r_{2} O_{x}$
$0.15 \times 2 = \frac{12.6}{104 + 16 x} \times 2 (6 - x)$
$x = 4$
Formula    $= C r_{2} o_{4}$ (or) $= C r O_{2}$

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