Cl2 gas is passed aqueous NaOH solution under given conditionsa Cl2 - NaOH cold - b Cl2 - NaOH hot - Equivalent mass of Cl2 in a and b is -

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Balancing Redox Equations

Cl2 gas is pa...

Question

$C l_{2}$ gas is passed aqueous NaOH solution under given conditions
(a) $C l_{2}$ + NaOH (cold) $\to$
(b) $C l_{2}$ + NaOH (hot) $\to$

Equivalent mass of $C l_{2}$ in (a) and (b) is :

Equivalent mass of

C l_{2}

in (a) is 71.0 g equivalent

^{- 1}

Equivalent mass of

C l_{2}

in (b) is 42.6 g equivalent

^{- 1}

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Equivalent mass of

C l_{2}

in (a) is 142.0 g equivalent

^{- 1}

Equivalent mass of

C l_{2}

in (b) is 85.2 g equivalent

^{- 1}

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Equivalent mass of

C l_{2}

in (a) is 142.0 g equivalent

^{- 1}

Equivalent mass of

C l_{2}

in (b) is 42.6 g equivalent

^{- 1}

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Equivalent mass of

C l_{2}

in (a) is 71.0 g equivalent

^{- 1}

Equivalent mass of

C l_{2}

in (b) is 85.2 g equivalent

^{- 1}

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None of the above

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Solution

The correct option is A Equivalent mass of $C l_{2}$ in (a) is 71.0 g equivalent $^{- 1}$
Equivalent mass of $C l_{2}$ in (b) is 42.6 g equivalent $^{- 1}$
(a) $C l_{2} + 2 N a O H \to N a C l + N a O C l + H_{2} O$

In hypochlorite ion NaOCl, the chlorine atom is in the +1 oxidation state. In NaCl, chlorine is at -1 oxidation state.

Equivalent mass = $\frac{m o l a r m a s s}{n - f a c t o r}$

n factor is = 1

Therefore, equivalent mass is $71.0$ g.

(b) $3 C l_{2} + 5 N a O H \to 5 N a C l + N a C l O_{3} + 3 H_{2} O$

In chlorite ion, the chlorine atom is in the +5 oxidation state.

In this, the n factor = 5

Therefore, equivalent mass is $\frac{71 \times 3}{5} = 42.6$ g

Hence, option A is correct.

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Similar questions

Q. Equivalent mass of

S

S {C l}_{2}

is 16. What is the equivalent mass of

S

S_{2} {C l}_{2}

C l_{2}

disproportionate to

C l^{-}

and species

A

, such that the equivalent mass of

C l_{2}

71

. Thus, the oxidation number of chlorine in species

A

is:

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

has the molar mass

M

, in the reaction

N a_{2} S_{2} O_{3} + H_{2} O + C l_{2} \to N a_{2} S O_{4} + 2 H C l + S

the equivalent weight of

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

is:

Q. The equivalent weights of

S

S {C l}_{2}

and

S_{2} {C l}_{2}

are in the ratio:

Q. Assertion :One equivalent of

M n O_{2}

reacts with 2 equivalents of

H C l

in the reaction,

M n O_{2} + 4 H C l ⟶ M n C l_{2} + 2 H_{2} O + C l_{2}

. Reason: One equivalent of

M n O_{2}

reacts with one equivalent of

H C l

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Cl2 gas is passed aqueous NaOH solution under given conditions(a) Cl2 + NaOH (cold) → (b) Cl2 + NaOH (hot) → Equivalent mass of Cl2 in (a) and (b) is :

$C l_{2}$ gas is passed aqueous NaOH solution under given conditions
(a) $C l_{2}$ + NaOH (cold) $\to$
(b) $C l_{2}$ + NaOH (hot) $\to$

Equivalent mass of $C l_{2}$ in (a) and (b) is :