For the coagulation of 200 mL of `As_(2)S_(3)` solution, 10 mL of 1 M NaCl is required. What is the coagulating value of NaCl.

Explanation

To find the coagulating value of NaCl for the coagulation of 200 mL of As₂S₃ solution, we can follow these steps:

### Step 1: Understand the concept of coagulating value
The coagulating value of an electrolyte is defined as the minimum concentration (in millimoles per liter) of the electrolyte required to cause the precipitation of a colloidal solution.

### Step 2: Calculate the amount of NaCl used
We are given that 10 mL of 1 M NaCl is required for the coagulation. To find the number of millimoles of NaCl used, we can use the formula:
\[ \text{Millimoles of NaCl} = \text{Volume (L)} \times \text{Molarity (mol/L)} \]
Converting 10 mL to liters:
\[ 10 \text{ mL} = 0.010 \text{ L} \]
Now, calculating the millimoles:
\[ \text{Millimoles of NaCl} = 0.010 \text{ L} \times 1 \text{ mol/L} = 0.010 \text{ mol} = 10 \text{ millimoles} \]

### Step 3: Relate the amount of NaCl to the volume of As₂S₃ solution
We know that 10 millimoles of NaCl are required to coagulate 200 mL of As₂S₃ solution. To find the coagulating value in terms of millimoles per liter, we need to scale this to 1 liter (1000 mL).

### Step 4: Calculate the coagulating value
To find the coagulating value of NaCl, we can set up a proportion:
\[ \text{Coagulating value} = \frac{\text{Millimoles of NaCl required for 200 mL}}{\text{Volume of As₂S₃ solution in L}} \]
Converting 200 mL to liters:
\[ 200 \text{ mL} = 0.200 \text{ L} \]
Now, substituting the values:
\[ \text{Coagulating value} = \frac{10 \text{ millimoles}}{0.200 \text{ L}} = 50 \text{ millimoles/L} \]

### Final Answer
The coagulating value of NaCl is **50 millimoles per liter**.

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