With 4 mole of Kl one mole of `Cl_(2)` is treated to yield gas, which is then treated with hypo solution, then correct regarding this is/are

Options

A. Equivalent weight of `I_(2) ` is `(M)/(1)`
B. To react completed with product 1 mole of hypo solution are required
C. n-factor of hypo solution is 1
D. n-factor of `Cl_(2)` is 2

Answer

To solve the problem step by step, we need to analyze the reaction between potassium iodide (KI) and chlorine gas (Cl₂) and the subsequent reaction with sodium thiosulfate (hypo solution).

### Step 1: Write the Reaction Between KI and Cl₂
When 4 moles of KI react with 1 mole of Cl₂, the reaction can be represented as follows:

\[ 2 \text{KI} + \text{Cl}_2 \rightarrow 2 \text{KCl} + \text{I}_2 \]

### Step 2: Balance the Reaction
To balance the reaction, we need to ensure that the number of atoms of each element is the same on both sides. The balanced equation is:

\[ 4 \text{KI} + \text{Cl}_2 \rightarrow 4 \text{KCl} + 2 \text{I}_2 \]

### Step 3: Determine the Products
From the balanced equation, we see that the products are potassium chloride (KCl) and iodine (I₂).

### Step 4: Reaction of I₂ with Hypo Solution
Next, we need to consider the reaction of iodine (I₂) with sodium thiosulfate (hypo solution, Na₂S₂O₃):

\[ \text{I}_2 + 2 \text{Na}_2\text{S}_2\text{O}_3 \rightarrow \text{Na}_2\text{S}_4\text{O}_6 + 2 \text{NaI} \]

### Step 5: Calculate the Amount of Hypo Solution Required
From the stoichiometry of the reaction, we see that 1 mole of I₂ reacts with 2 moles of Na₂S₂O₃. Since we produced 1 mole of I₂ from the reaction with Cl₂, we will need:

\[ 2 \text{ moles of Na}_2\text{S}_2\text{O}_3 \]

### Step 6: Determine the N-Factor of I₂
The n-factor of a substance is defined as the number of moles of electrons transferred per mole of the substance in a reaction. For I₂, the change in oxidation state from -1 in KI to 0 in I₂ indicates that each iodine atom undergoes a change of 1. Since there are 2 iodine atoms in I₂, the n-factor for I₂ is:

\[ \text{n-factor of I}_2 = 2 \]

### Step 7: Determine the N-Factor of Hypo Solution
To find the n-factor of sodium thiosulfate (Na₂S₂O₃), we need to analyze its reaction with iodine. The oxidation state of sulfur in Na₂S₂O₃ changes during the reaction. The n-factor is determined based on the number of moles of electrons transferred. In this case, the n-factor is found to be 1.

### Step 8: Determine the N-Factor of Cl₂
For chlorine (Cl₂), the oxidation state changes from 0 in Cl₂ to -1 in KCl. Since there are 2 chlorine atoms, the n-factor for Cl₂ is:

\[ \text{n-factor of Cl}_2 = 2 \]

### Conclusion
Based on the above analysis, the correct statements regarding the reactions are:
1. The equivalent weight of I₂ is m/2 (incorrect).
2. To react completely with the product, 2 moles of hypo solution are required (incorrect).
3. The n-factor of hypo solution is 1 (correct).
4. The n-factor of Cl₂ is 2 (correct).