Gold number of haemoglobin is 0.03. Hence 100 ml of gold sol will require haemoglobin so that gold is not coagulated by 10 ml of 10% NaCl solution

Options

A. 0.03 mg
B. 30 mg
C. 0.30 mg
D. 3 mg

Answer

To solve the problem, we need to determine the amount of hemoglobin required to prevent the coagulation of 100 ml of gold sol when 10 ml of 10% NaCl solution is added. We are given that the gold number of hemoglobin is 0.03.

### Step-by-Step Solution:

1. **Understand the Gold Number**:
The gold number of a protective colloid (in this case, hemoglobin) is defined as the minimum amount (in mg) of the protective colloid required to prevent the coagulation of 10 ml of gold sol when 1 ml of a standard coagulating agent (like NaCl) is added.

2. **Identify the Given Values**:
- Gold number of hemoglobin = 0.03 mg for 10 ml of gold sol.
- Volume of gold sol = 100 ml.
- Volume of NaCl solution = 10 ml.

3. **Set Up the Proportion**:
Since the gold number is given for 10 ml of gold sol, we can set up a proportion to find the amount of hemoglobin required for 100 ml of gold sol.
\[ \text{If 0.03 mg is needed for 10 ml, then } x \text{ mg is needed for 100 ml.} \]

4. **Use Cross-Multiplication**:
\[ x = \frac{100 \, \text{ml} \times 0.03 \, \text{mg}}{10 \, \text{ml}} \]
Simplifying this gives:
\[ x = \frac{3 \, \text{mg}}{1} = 0.30 \, \text{mg} \]

5. **Conclusion**:
The amount of hemoglobin required to prevent the coagulation of 100 ml of gold sol when 10 ml of 10% NaCl solution is added is **0.30 mg**.

### Final Answer:
The correct option is **0.30 mg**.

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