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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85.17

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Solution

The correct option is D 85.17
$14 1 + H 1 - C l \to 1 + K 1 - C l + 2 3 + C r {1 - C l}_{3} + 3 {0 C l}_{2} + {1 + H}_{2} 2 - O$
$n_{f} = \frac{6 (0 - (- 1))}{14} = 6 / 14$
$E q l^{t} w e i g h t = \frac{m o l e c u l a r w e i g h t}{n_{f}} = 85.17$

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