5 ml `As_(2)S_(3)` is mixed with distilled water and 0.01 M solution of an electrolyte AB so that total volume is 10 ml. It was found that all solution containing more than 5 ml of AB coagulate within 5 min . What is the Flocculation value of AB for `As_(2)S_(3)`sol?

Answer

To solve the problem, we need to find the flocculation value of the electrolyte AB for the colloid As₂S₃. Here’s a step-by-step breakdown of the solution:

### Step 1: Understanding the Problem
We have a 5 ml solution of As₂S₃ mixed with distilled water and a 0.01 M solution of an electrolyte AB to make a total volume of 10 ml. We need to determine the flocculation value of AB for As₂S₃.

### Step 2: Determine the Volume of AB Added
Since the total volume is 10 ml and we have 5 ml of As₂S₃, the volume of the electrolyte AB added must also be 5 ml (10 ml - 5 ml = 5 ml).

### Step 3: Calculate the Initial Millimoles of AB
To find the initial millimoles of AB in the 5 ml solution:
- Molarity (M) = 0.01 M
- Volume (V) = 5 ml = 0.005 L

Using the formula:
\[ \text{Millimoles} = \text{Molarity} \times \text{Volume (in L)} = 0.01 \, \text{mol/L} \times 0.005 \, \text{L} = 0.00005 \, \text{mol} = 0.05 \, \text{mmol} \]

### Step 4: Understanding Coagulation
It is given that solutions containing more than 5 ml of AB coagulate within 5 minutes. This indicates that 5 ml of AB is the minimum amount required for coagulation.

### Step 5: Calculate the Flocculation Value
The flocculation value is defined as the minimum millimole of electrolyte required to coagulate 1000 ml of colloidal solution. Since we have determined that 0.05 mmol of AB is sufficient to coagulate 10 ml of As₂S₃, we can scale this up to find the amount needed for 1000 ml.

Using the proportion:
\[ \text{Flocculation Value} = \left(\frac{0.05 \, \text{mmol}}{10 \, \text{ml}}\right) \times 1000 \, \text{ml} = 5 \, \text{mmol} \]

### Conclusion
The flocculation value of AB for As₂S₃ is **5 mmol**.

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