Equivalent weight of I 2 in the reaction, I 2+5 Cl 2+6 H 2 O → 2 HIO 3+10 HCl, is

Question

Equivalent weight of I 2 in the reaction, I 2+5 Cl 2+6 H 2 O → 2 HIO 3+10 HCl, is (A) M (B) M / 2 (C) M / 5 (D) M / 10

Tardigrade Admin · Accepted Answer

overset0I2+5Cl2+6H2O → 2 overset+5HIO3+10HCl Change in oxidation number = 5 Equivalent weight =( text Atomic weight / text Change in oxidation number )=(M/5)

Q. Equivalent weight of $I_{2}$ in the reaction, $I_{2} + 5 C l_{2} + 6 H_{2} O \to 2 H I O_{3} + 10 H Cl$ , is

$M$

$M /2$

$M /5$

$M /10$

$I_{2} 0 + 5 C l_{2} + 6 H_{2} O \to 2 H I O_{3} + 5 + 10 H Cl$

Change in oxidation number = $5$

Equivalent weight $= \frac{Atomic weight}{Change in oxidation number} = \frac{M}{5}$

Q. Equivalent weight of I2​ in the reaction, I2​+5Cl2​+6H2​O→2HIO3​+10HCl, is

M

M/2

M/5

M/10

I2​0​+5Cl2​+6H2​O→2HIO3​+5​+10HCl Change in oxidation number = 5 Equivalent weight = Change in oxidation number Atomic weight ​=5M​

Q. Equivalent weight of $I_{2}$ in the reaction, $I_{2} + 5 C l_{2} + 6 H_{2} O \to 2 H I O_{3} + 10 H Cl$ , is

$M$

$M /2$

$M /5$

$M /10$

$I_{2} 0 + 5 C l_{2} + 6 H_{2} O \to 2 H I O_{3} + 5 + 10 H Cl$

Change in oxidation number = $5$

Equivalent weight $= \frac{Atomic weight}{Change in oxidation number} = \frac{M}{5}$