In a coagulation experiment, `5mL` of `As_(2)S_(3)` is mixed with distilled water and `0.1M` solution of an electrolyte `AB` so that the total volume is `10mL` . It was found that all solutions containing more than `4.6mL` . Of `AB` coagulate within `5` min. What is the flocculation value of `AB` for `As_(2)S_(3)` solution?

Explanation

To find the flocculation value of the electrolyte \( AB \) for the \( As_2S_3 \) solution, we can follow these steps:

### Step 1: Understand the given data
- Volume of \( As_2S_3 \) solution = \( 5 \, \text{mL} \)
- Volume of distilled water = \( 10 \, \text{mL} - 5 \, \text{mL} = 5 \, \text{mL} \)
- Concentration of \( AB \) = \( 0.1 \, \text{M} \)
- Total volume of the solution = \( 10 \, \text{mL} \)
- It is found that solutions containing more than \( 4.6 \, \text{mL} \) of \( AB \) coagulate within \( 5 \, \text{min} \).

### Step 2: Calculate the volume of \( AB \) required for coagulation
Since it is stated that more than \( 4.6 \, \text{mL} \) of \( AB \) is required for coagulation, we can take \( 5 \, \text{mL} \) as the minimum volume required for coagulation.

### Step 3: Calculate the moles of \( AB \) in the solution
Using the formula for moles:
\[ \text{Moles of } AB = \text{Volume (L)} \times \text{Concentration (M)} \]
Convert \( 5 \, \text{mL} \) to liters:
\[ 5 \, \text{mL} = 0.005 \, \text{L} \]
Now, calculate the moles of \( AB \):
\[ \text{Moles of } AB = 0.005 \, \text{L} \times 0.1 \, \text{M} = 0.0005 \, \text{moles} \]

### Step 4: Convert moles to millimoles
Since \( 1 \, \text{mole} = 1000 \, \text{millimoles} \):
\[ 0.0005 \, \text{moles} = 0.5 \, \text{millimoles} \]

### Step 5: Calculate the flocculation value
The flocculation value is defined as the minimum amount of electrolyte required for coagulation of \( 1 \, \text{L} \) of solution. Since \( 0.5 \, \text{millimoles} \) of \( AB \) is required for \( 10 \, \text{mL} \) of solution, we can calculate the amount required for \( 1 \, \text{L} \):
\[ \text{Flocculation value} = 0.5 \, \text{millimoles} \times 100 = 5 \, \text{millimoles} \]

### Final Answer
The flocculation value of \( AB \) for \( As_2S_3 \) solution is \( 5 \, \text{millimoles} \).

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