${3 C H}_{3} {C H}_{2} O H + 2 K_{2} {C r}_{2} O_{7} + 8 H_{2} S O_{4} ⟶ 3 {C H}_{3} C O O H + 2 {C r}_{2} {(S O_{4})}_{3}$
$+ {2 K}_{2} S O_{4} + 11 H_{2} O$

Equivalent mass of ${C H}_{3} {C H}_{2} O H$ in the reaction is :

\frac{m o l a r m a s s}{1}

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\frac{m o l a r m a s s}{2}

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\frac{m o l a r m a s s}{3}

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\frac{m o l a r m a s s}{4}

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Solution

The correct option is B $\frac{m o l a r m a s s}{2}$
In the above reaction,

$C H_{3} C H_{2} O H$ (alcohol) is oxidized into $C H_{3} C O O H$ (carboxylic acid)

the electron change involve in this conversion is 2.

so, n-fator becomes 2,

we have, equivalent mass = $\frac{m o l a r m a s s}{n - f a c t o r}$

equivalent mass of $C H_{3} C H_{2} O H$ = $\frac{m o l a r m a s s}{2}$
Option B is correct

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Similar questions

F e S_{2} + K M n O_{4} + H^{+} \to F e^{+ 3} + S O_{2} + M n^{+ 2} + H_{2} O

F e S_{2}

{3 C H}_{3} {C H}_{2} O H + 2 K_{2} {C r}_{2} O_{7} + 8 H_{2} S O_{4} ⟶ 3 {C H}_{3} C O O H + 2 {C r}_{2} {(S O_{4})}_{3}

+ {2 K}_{2} S O_{4} + 11 H_{2} O

K_{2} {C r}_{2} O_{7}

H_{3} C C H_{2} O H + 2 K_{2} C r_{2} O_{7} + 8 H_{2} S O_{4} \to

3 C H_{3} C O O H + 2 C r_{2} (S O_{4})_{3} + 2 K_{2} S O_{4} + 11 H_{2} O

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

X e O_{3} + 6 H^{+} + 6 F e^{2 +} ⟶ X e + 3 H_{2} O + 6 F e^{2 +}

e q u i v a l e n t m a s s o f X e O_{3} = \frac{m o l a r m a s s}{. . . . . . . . . .}

X e F_{2} + {B r O_{3}}^{-} + H_{2} O ⟶ X e + {B r O_{4}}^{-} + 2 H F

e q u i v a l e n t m a s s o f {B r O_{3}}^{-} = \frac{m o l a r m a s s}{. . . . . . . . . .}

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