Question

Calculate the quantity of fertilizer required for making a 100 ppm solution of 20-10-20 in a 500 litre tank?

• A. 150 g
• B. 250 g
• C. 350 g
• D. 00 g

Hint:

To quickly find the fertilizer amount in grams for a given ppm of N:
\[ \text{Fertilizer (g)} = \frac{\text{ppm} \times \text{Volume (L)}}{\% \text{ Nutrient} \times 10} \] Here, \[ \frac{100 \times 500}{20 \times 10} = \frac{50000}{200} = 250 \text{ g} \]

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
Parts per million (ppm) is a unit of concentration defined as milligrams of solute per liter of solvent ($\text{mg/L}$).
When a fertilizer grade like 20-10-20 is specified, the numbers represent the percentages of Nitrogen ($\text{N}$), Phosphorus ($\text{P}_2\text{O}_5$), and Potassium ($\text{K}_2\text{O}$) in the fertilizer, respectively.
A "100 ppm solution of 20-10-20" usually implies delivering 100 ppm of the primary nutrient, Nitrogen ($\text{N}$), using this fertilizer source.
Key Formula or Approach:
The mass of nutrient required to achieve a target concentration is: \[ \text{Mass of Nutrient (mg)} = \text{Target Concentration (ppm)} \times \text{Volume of Water (L)} \] Since the fertilizer contains only a fraction of the target nutrient, the quantity of fertilizer is: \[ \text{Quantity of Fertilizer} = \frac{\text{Mass of Nutrient}}{\text{Nutrient Percentage in Fertilizer}} \]

Step 2: Detailed Explanation:
Calculate the total mass of Nitrogen required for 500 liters at 100 ppm: \[ \text{Mass of N} = 100 \text{ mg/L} \times 500 \text{ L} = 50,000 \text{ mg} = 50 \text{ g} \] The fertilizer grade is 20-10-20, meaning it contains 20% Nitrogen by weight.
Calculate the required mass of the fertilizer: \[ \text{Quantity of Fertilizer} = \frac{50 \text{ g}}{20\%} = \frac{50}{0.20} = 250 \text{ g} \]

Step 3: Final Answer:
The quantity of fertilizer required is 250 g.