The coagulation of 100ml of colloidal solution of gold is completely prevented by addition of `0.25 g` of a substance "X" to it before addition of 1 ml of `10%` NaCI solution. The gold number of "X"is :

Options

A. 0.25
B. 25
C. 250
D. 2.5

Explanation

To find the gold number of the substance "X" that prevents the coagulation of a colloidal gold solution, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Problem**:
- We have a colloidal solution of gold (100 mL).
- The coagulation of this solution is prevented by adding 0.25 g of substance "X" before adding 1 mL of 10% NaCl solution.

2. **Convert the Mass of Substance X to Milligrams**:
- Since the gold number is expressed in milligrams, we need to convert grams to milligrams.
- \( 0.25 \text{ g} = 0.25 \times 1000 \text{ mg} = 250 \text{ mg} \)

3. **Identify the Volume of the Coagulating Agent**:
- The volume of the NaCl solution added is 1 mL of a 10% solution.
- The concentration of NaCl does not directly affect the calculation of the gold number, but it indicates that the coagulation is being tested.

4. **Apply the Formula for Gold Number**:
- The gold number is defined as the number of milligrams of a substance that prevents the coagulation of 10 mL of a colloidal gold solution when a specific amount of an electrolyte (in this case, NaCl) is added.
- Since we are dealing with 100 mL of colloidal solution, we need to adjust the gold number accordingly. The formula can be simplified as:
\[ \text{Gold Number (GN)} = \frac{\text{mass of substance X (in mg)}}{\text{volume of colloidal solution (in mL)}} \]

5. **Calculate the Gold Number**:
- Here, the mass of substance X is 250 mg, and the volume of the colloidal solution is 100 mL.
- Therefore, the gold number is:
\[ \text{GN} = \frac{250 \text{ mg}}{100 \text{ mL}} = 2.5 \text{ mg/mL} \]

6. **Final Result**:
- The gold number of substance "X" is 2.5.

### Conclusion:
The gold number of substance "X" is **2.5**.