In the following reaction- Na2S2O3 is converted to Na2S4O6-2Na2S2O3 -I2- Na2S4O6 - 2NaIThe equivalent weight of Na2S2O3 for this reaction is-The molecular weight of Na2S2O3 - M

The molecular weight of Na_2S_2O_3 = M Na_2S_2O_3 nbsp; + 12I_2→12 Na_2S_4O_6 + NaI Oxidation state of S nbsp; in nbsp; Na_2S_2O_3 = + 2 Oxidation ...

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Byju's Answer

Standard XII

Chemistry

Equivalent Mass

In the follow...

Question

In the following reaction, $N a_{2} S_{2} O_{3}$ is converted to $N a_{2} S_{4} O_{6}$ .
$2 N a_{2} S_{2} O_{3} + I_{2} \to N a_{2} S_{4} O_{6} + 2 N a I$
The equivalent weight of $N a_{2} S_{2} O_{3}$ for this reaction is:
(The molecular weight of $N a_{2} S_{2} O_{3} = M)$

M

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\frac{M}{4}

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\frac{M}{2}

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\frac{M}{3}

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Solution

The correct option is A $M$
The molecular weight of $N a_{2} S_{2} O_{3} = M N a_{2} S_{2} O_{3} + \frac{1}{2} I_{2} \to \frac{1}{2} N a_{2} S_{4} O_{6} + N a I Oxidation state of S in N a_{2} S_{2} O_{3} = + 2 Oxidation state of S in N a_{2} S_{4} O_{6} = + 2.5 ∴ Δ O . S . = 2.5 - 2 = 0.5 ∴ n_{f} = 2 \times 0.5 = 1 Where, O.S. is the oxidation state Equivalent weight = \frac{Molecular weight}{n_{f}} = \frac{M}{1} = M$

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Similar questions

Q. In the following reaction,

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

is converted to

N a_{2} S_{4} O_{6}

2 N a_{2} S_{2} O_{3} + I_{2} \to N a_{2} S_{4} O_{6} + 2 N a I

The equivalent weight of

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

for this reaction is:

(The molecular weight of

N a_{2} S_{2} O_{3} = M)

Q. In the reaction:

N a_{2} S_{2} O_{3} + 4 C l_{2} + 5 H_{2} O \to N a_{2} S O_{4} + 8 H C l

, the equivalent weight of

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

will be (M

=

molecular weight of

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

Q. The equivalent weight of

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

(Mol. wt . = M ) in the reaction,

2 N a_{2} S_{2} O_{3} + I_{2} \to N a_{2} S_{4} O_{6} + 2 N a I

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. In the reaction,

2 S_{2} O_{3}^{2 -} + I_{2} ⟶ S_{4} O_{6}^{2 -} + 2 I^{-}

, the equivalent weight of

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

is equal to its.

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In the following reaction, Na2S2O3 is converted to Na2S4O6. 2Na2S2O3+I2→Na2S4O6 + 2NaI The equivalent weight of Na2S2O3 for this reaction is: (The molecular weight of Na2S2O3=M)

The correct option is A MThe molecular weight of Na2S2O3=MNa2S2O3 + 12I2→12Na2S4O6 + NaIOxidation state of S in Na2S2O3=+2Oxidation state of S in Na2S4O6=+2.5∴ΔO.S.=2.5−2=0.5∴nf=2×0.5=1Where, O.S. is the oxidation stateEquivalent weight=Molecular weightnf =M1=M

In the following reaction, $N a_{2} S_{2} O_{3}$ is converted to $N a_{2} S_{4} O_{6}$ .
$2 N a_{2} S_{2} O_{3} + I_{2} \to N a_{2} S_{4} O_{6} + 2 N a I$
The equivalent weight of $N a_{2} S_{2} O_{3}$ for this reaction is:
(The molecular weight of $N a_{2} S_{2} O_{3} = M)$