In an experiment, addition of `5.0mL` , of `0.006MBaCl_(2)` to `10.0mL` of arsenic sulphite sol just causes the complete coagulation in `34h` . The flocculating value of the effective ion is:

Options

A. 2
B. 3
C. 4
D. 5

Explanation

To find the flocculating value of the effective ion in the given experiment, we can follow these steps:

### Step 1: Identify the given data
- Volume of BaCl₂ solution added = 5.0 mL
- Molarity of BaCl₂ solution = 0.006 M
- Volume of arsenic sulphide sol = 10.0 mL
- Time for complete coagulation = 34 hours (not needed for calculation)

### Step 2: Calculate the total volume of the solution
The total volume after adding BaCl₂ to the arsenic sulphide sol is:
\[ \text{Total Volume} = \text{Volume of BaCl₂} + \text{Volume of Arsenic Sulphide Sol} = 5.0 \, \text{mL} + 10.0 \, \text{mL} = 15.0 \, \text{mL} \]

### Step 3: Convert the volume of BaCl₂ to liters
To convert milliliters to liters, we divide by 1000:
\[ \text{Volume of BaCl₂ in liters} = \frac{5.0 \, \text{mL}}{1000} = 0.005 \, \text{L} \]

### Step 4: Calculate the number of moles of BaCl₂
Using the formula:
\[ \text{Moles} = \text{Volume (L)} \times \text{Molarity} \]
Substituting the values:
\[ \text{Moles of BaCl₂} = 0.005 \, \text{L} \times 0.006 \, \text{M} = 0.00003 \, \text{moles} \]

### Step 5: Convert moles of BaCl₂ to millimoles
Since 1 mole = 1000 millimoles:
\[ \text{Millimoles of BaCl₂} = 0.00003 \, \text{moles} \times 1000 = 0.03 \, \text{mmol} \]

### Step 6: Determine the flocculating value
The flocculating value is defined as the minimum amount of electrolyte (in millimoles) required for the complete coagulation of 1 liter of colloidal solution. Since we have the total volume of the solution as 15 mL, we need to scale up the millimoles calculated to 1 liter (1000 mL):
\[ \text{Flocculating Value} = \frac{0.03 \, \text{mmol}}{0.015 \, \text{L}} \times 1 \, \text{L} = 2 \, \text{mmol} \]

### Conclusion
The flocculating value of the effective ion (Ba²⁺) is **2 mmol/L**.

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