For coagulation of 195 ml of `As_(2)S_(3)` solution 10 ml of 1 M NaCl is required. Calculate the coagulating power of `NaCl`.

Answer

To calculate the coagulating power of NaCl for the coagulation of As₂S₃ solution, we will follow these steps:

### Step 1: Understand the Given Information
We have:
- Volume of As₂S₃ solution = 195 ml
- Volume of NaCl solution used = 10 ml
- Concentration of NaCl = 1 M

### Step 2: Calculate the Millimoles of NaCl Used
To find the number of millimoles of NaCl used, we use the formula:
\[ \text{Millimoles of NaCl} = \text{Volume (ml)} \times \text{Concentration (M)} \]
Substituting the values:
\[ \text{Millimoles of NaCl} = 10 \, \text{ml} \times 1 \, \text{mmol/ml} = 10 \, \text{mmol} \]

### Step 3: Calculate the Coagulating Power
Coagulating power is defined as the number of millimoles of the electrolyte required to coagulate 1 liter of the colloidal solution.

First, we find out how many millimoles of NaCl are required to coagulate 1 ml of As₂S₃ solution:
\[ \text{Millimoles of NaCl per ml of As₂S₃} = \frac{10 \, \text{mmol}}{195 \, \text{ml}} = \frac{10}{195} \approx 0.05128 \, \text{mmol/ml} \]

Now, to find out how many millimoles are needed for 1 liter (1000 ml) of As₂S₃ solution:
\[ \text{Millimoles of NaCl for 1 liter of As₂S₃} = 0.05128 \, \text{mmol/ml} \times 1000 \, \text{ml} = 51.28 \, \text{mmol} \]

### Step 4: Conclusion
The coagulating power of NaCl is therefore:
\[ \text{Coagulating Power of NaCl} = 51.28 \, \text{mmol/L} \]

### Final Answer
The coagulating power of NaCl is **51.28 mmol/L**.

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