Question

For preparing 200 ppm solution of IAA, the amount of IAA required is

• A. 2 mg
• B. 20 mg
• C. 200 mg
• D. 2000 mg

Hint:

Remember ppm calculations:
- 1 ppm = 1 mg/L
- 200 ppm = 200 mg/L
- For 1 L: 200 mg
- For 100 mL: 20 mg
- For 50 mL: 10 mg
Common plant growth regulators: IAA, IBA, NAA, GA3, Kinetin.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
This question tests knowledge of calculating concentrations of plant growth regulators.

Step 2: Key Formula or Approach:
ppm = mg/L (parts per million).
1 ppm = 1 mg/L.
For preparing 200 ppm solution:
Need 200 mg of solute per 1 L of solution.
For 1 L of solution, 200 mg is required.
But if we are preparing a smaller volume (e.g., 100 mL), the amount would be 20 mg.
The question asks for the amount required (assuming 100 mL, which is common for laboratory solutions).

Step 3: Detailed Explanation:
Given: 200 ppm solution of IAA.
If we need to prepare 1 L (1000 mL) of 200 ppm solution:
\[ \text{IAA required} = 200 \text{ mg} \] If we need to prepare 100 mL (0.1 L) of 200 ppm solution:
\[ \text{IAA required} = 200 \text{ mg/L} \times 0.1 \text{ L} = 20 \text{ mg} \] Since the question asks for "the amount of IAA required" (for a standard preparation),
the answer is 20 mg (for 100 mL solution).

Step 4: Final Answer:
Thus, the amount of IAA required is 20 mg, which corresponds to option (B).
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