Question:

The equivalent weight of Na2S2O3\text{Na}_2\text{S}_2\text{O}_3 (Gram molecular weight = MM) in the given reaction is:
I2+2Na2S2O32NaI+Na2S4O6\text{I}_2 + 2 \text{Na}_2\text{S}_2\text{O}_3 \rightarrow 2 \text{NaI} + \text{Na}_2\text{S}_4\text{O}_6

Updated On: Apr 3, 2025
  • M/2
  • M
  • 2M
  • M/4
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The Correct Option is B

Approach Solution - 1

The reaction involves: I2+2Na2S2O32NaI+Na2S4O6 \text{I}_2 + 2\text{Na}_2\text{S}_2\text{O}_3 \rightarrow 2\text{NaI} + \text{Na}_2\text{S}_4\text{O}_6 .

Here, each molecule of Na2S2O3\text{Na}_2\text{S}_2\text{O}_3 donates 1 electron, leading to a total of 2 electrons being transferred per molecule of I2\text{I}_2.

Oxidation states of sulfur: In Na2S2O3\text{Na}_2\text{S}_2\text{O}_3, the average oxidation state of sulfur is +2. During the reaction, sulfur in Na2S2O3\text{Na}_2\text{S}_2\text{O}_3 is partially oxidized to Na2S4O6\text{Na}_2\text{S}_4\text{O}_6, where the average oxidation state becomes +2.5.

Definition of equivalent weight: The equivalent weight of a substance is given by: Equivalent Weight=Molecular Weightn. \text{Equivalent Weight} = \frac{\text{Molecular Weight}}{n}. Here, n represents the number of electrons transferred per molecule in the reaction.

Calculation: For Na2S2O3\text{Na}_2\text{S}_2\text{O}_3, n=1n = 1, because each molecule transfers 1 electron. Therefore, the equivalent weight of Na2S2O3\text{Na}_2\text{S}_2\text{O}_3 is equal to its molecular weight, M. Thus, the equivalent weight of Na2S2O3\text{Na}_2\text{S}_2\text{O}_3 in this reaction is M.

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Approach Solution -2

Correct Answer:

Option 2: M

Explanation:

1. Analyze the Reaction:

I2 + 2Na2S2O3 → 2NaI + Na2S4O6

2. Determine the Change in Oxidation State:

In Na2S2O3, the average oxidation state of sulfur is +2.

In Na2S4O6, the average oxidation state of sulfur is +2.5.

3. Calculate the Change per Molecule:

Each S atom in Na2S2O3 changes its oxidation state from +2 to +2.5, a change of 0.5.

There are two sulfur atoms in Na2S2O3, so the total change in oxidation state per molecule of Na2S2O3 is 2 * 0.5 = 1.

4. Determine the n-factor:

The n-factor (number of electrons transferred per molecule) is equal to the total change in oxidation state, which is 1.

5. Calculate the Equivalent Weight:

Equivalent weight = Molecular weight / n-factor

Equivalent weight = M / 1 = M

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