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. 500 g

Hint:

Steps for ppm calculation: 1. ppm means "parts per million" or mg/L. So, 100 ppm N = 100 mg N per liter. 2. Calculate total N needed: 100 mg/L * 500 L = 50,000 mg N = 50 g N. 3. Fertilizer is 20% N. This means 0.20 g N per 1 g fertilizer. 4. Amount of fertilizer = (Total N needed) / (N_concentration_in_fertilizer) = 50 g / 0.20 = 250 g.

Answer 1

The Correct Option is B

The question asks for a 100 ppm solution using a 20-10-20 fertilizer. When a ppm solution is specified without mentioning which nutrient, it usually refers to ppm of the nutrient that is present in the highest percentage, or ppm of Nitrogen (N) if it's a complete fertilizer and N is of primary interest for growth stimulation. Here, Nitrogen (N) is 20% and Potassium (K\(_2\)O) is 20%. Let's assume the 100 ppm refers to Nitrogen (N). 1 ppm = 1 mg of solute per liter of water. So, 100 ppm N = 100 mg of N per liter of water. Total Nitrogen needed for 500 liters of water: Total N = 100 mg/L \( \times \) 500 L = 50,000 mg of N. Convert mg to grams: 50,000 mg = 50 g of N. The fertilizer grade is 20-10-20, which means: It contains 20% Nitrogen (N) by weight. This means 100 g of fertilizer contains 20 g of N. Or, to get 20 g of N, we need 100 g of fertilizer. To get 1 g of N, we need \( \frac10020 \) g of fertilizer = 5 g of fertilizer. We need 50 g of N. Quantity of fertilizer required = Amount of N needed \( \times \) (Amount of fertilizer per unit of N) Quantity of fertilizer required = 50 g of N \( \times \) 5 g fertilizer/g N Quantity of fertilizer required = 250 g. Alternatively, let X be the grams of fertilizer needed. 20% of X must be equal to 50 g of N. 0.20 \( \times \) X = 50 g X = \( \frac500.20 \) g X = \( \frac5020/100 \) g = \( \frac50 \times 10020 \) g = \( \frac500020 \) g = 250 g. Therefore, 250 g of 20-10-20 fertilizer is required. 250 g