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Electrolysis and Electrolytes

In the reacti...

Question

In the reaction,
$2 C u S O_{4} + 4 K I ⟶ C u_{2} I_{2} + 2 K_{2} S O_{4} + I_{2}$ , the ratio of equivalent mass of $C u S O_{4}$ to its molar mass is :

\frac{1}{8}

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\frac{1}{4}

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\frac{1}{2}

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Solution

The correct option is D 1
$2 C u S O_{4} + 4 K I ⟶ C u_{2} I_{2} + 2 K_{2} S O_{4} + I_{2}$
+2 +1
So, Eq. mass of $C u S O_{4}$ is $M$ because, change in the oxidation state is 1.

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Similar questions

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

C u S O_{4}

(M w = 159.5 g

m o l^{- 1})

C u S O_{4} + 4 N H_{3} \to C u (N H_{3})_{4}] S O_{4}

N a_{2} S O_{4} + B a C l_{2} \to B a S O_{4} + 2 N a C l

S O_{2} + H_{2} O \to H_{2} S O_{3}

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

2 C u S O_{4} - ------- - + 4 K I ⟶ {C u}_{2} I_{2} + 2 K_{2} S O_{4} + I_{2}

M

C u S O_{4}

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

(a)

C u S O_{4}

K I

I_{2}

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

(b)

H g_{5} (I O_{6})_{2}

K I

H C l

H g_{5} (I O_{6})_{2} + 34 K I + 24 H C l ⟶ 5 K_{2} H g I_{4} + 8 I_{2} + 24 K C l + 12 H_{2} O

(c)

N a_{2} S_{2} O_{3}

0.0499

C u S O_{4} .5 H_{2} O

N a_{2} S_{2} O_{3}

I_{2}

0.7245

H g_{5} (I O_{6})_{2}

C u S O_{4} .5 H_{2} O = 249.5

H g_{5} (I O_{6})_{2} = 1448.5

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