Free iodine is titrated against standard reducing agent usually with sodium thiosulphate , i.e.
`K_(2)Cr_(2)O_(7) + 6Kl + 7H_(2)SO_(4) to Cr_(2)(SO_(4))_(3) + 4K_(2) SO_(4) + 7H_(2)O+l_(2)`
` 2CuSO_(4) + 4Kl to Cu_(2)l_(2) + 2K_(2)SO_(4) + l_(2)`
` l_(2) + Na_(2)S_(2)O_(3) to 2Nal + Na_(2)S_(4)O_(6)`
In iodometric titration, starch solution is used as an indicator. Starch solution gives blue or violet colour with free iodine .At the end point , blue or violet colour disappears when iodine is completely changed to iodide.
25 ml of N `K_(2)Cr_(2)O_(7)` acidified solution will liberate ............ iodine from Kl Solution :

Options

A. `0.3175` g
B. `3.175` g
C. `1.75` g
D. `317.5` g

Answer

To solve the problem, we need to determine how much iodine (I₂) is liberated from a potassium iodide (KI) solution when 25 mL of a normal potassium dichromate (K₂Cr₂O₇) solution is used in the titration. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the Reaction
The reaction between potassium dichromate and potassium iodide liberates iodine. The balanced equation shows that 1 mole of K₂Cr₂O₇ reacts with 6 moles of KI to produce 3 moles of I₂.

### Step 2: Calculate the Milliequivalents of K₂Cr₂O₇
We know that the normality (N) of K₂Cr₂O₇ is 1 N, and the volume (V) is 25 mL. The formula for calculating milliequivalents (meq) is:
\[ \text{Milliequivalents} = \text{Normality} \times \text{Volume in mL} \]
Substituting the values:
\[ \text{Milliequivalents of K₂Cr₂O₇} = 1 \, \text{N} \times 25 \, \text{mL} = 25 \, \text{meq} \]

### Step 3: Relate Milliequivalents of K₂Cr₂O₇ to I₂
From the stoichiometry of the reaction, we see that 1 equivalent of K₂Cr₂O₇ liberates 6 equivalents of I₂. Therefore, the milliequivalents of I₂ liberated will also be 25 meq (as they are equivalent):
\[ \text{Milliequivalents of I₂} = 25 \, \text{meq} \]

### Step 4: Calculate the Mass of I₂
To find the mass of iodine liberated, we use the formula:
\[ \text{Mass} = \text{Milliequivalents} \times \text{Equivalent weight} \times \frac{1}{1000} \]
The equivalent weight of iodine (I₂) can be calculated as follows:
- The molar mass of I₂ = 127 g/mol × 2 = 254 g/mol
- The equivalent weight of I₂ = Molar mass / 2 = 254 g/mol / 2 = 127 g/equiv

Now substituting the values:
\[ \text{Mass of I₂} = 25 \, \text{meq} \times 127 \, \text{g/equiv} \times \frac{1}{1000} \]
\[ = \frac{25 \times 127}{1000} = 3.175 \, \text{g} \]

### Conclusion
The mass of iodine liberated from the KI solution when 25 mL of 1 N K₂Cr₂O₇ is used is **3.175 g**.