The equivalent weight of $H C l$ in the given reaction is :
$K_{2} C r_{2} O_{7} + 14 H C l \to 2 K C l + 2 C r C l_{3} + 3 C l_{2} + H_{2} O$

16.25

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36.5

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85.1

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Solution

The correct option is B 85.1
In the given reaction, $H C l$ is oxidized to $C l_{2}$

Change in oxidation number $= 0 - (- 1) = 1$

$n f a c t o r = \frac{6}{14}$

$E q . w t . = \frac{M o l e c u l a r m a s s o f H C l}{n f a c t o r} = \frac{36.5}{\frac{6}{14}} = 85.1$

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Similar questions

H C l

K_{2} C r_{2} O_{7} + 14 H C l \to 2 K C l + 2 C r C l_{3} + 3 C l_{2} + H_{2} O

H C l

K_{2} C r_{2} O_{7} + 14 H C l \to 2 K C l + 2 C r C l_{3} + 3 C l_{2} + H_{2} O

n

H C l

K_{2} C r_{2} O_{7} + 14 H C l \to 2 K C l + 2 C r C l_{3} + 3 C l_{2} + 7 H_{2} O

K_{2} C r_{2} O_{7} + 14 H C l \to 2 K C l + 2 C r C l_{3} + 3 C l_{2} + 7 H_{2} O

H C l

M_{H C l} = 36.6 {g mol}^{- 1}

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