MP Board · Class 9 · Science · Improvement in Food ResourcesSolve the numerical problem in Part A and answer the theoretical question in Part B: Part A (Numerical): A farmer owns a crop field of area $3\text{ hectares}$. The required dosage of Nitrogen for his crop is $100\text{ kg}$ per hectare. He decides to fulfill this Nitrogen requirement using Ammonium Nitrate fertilizer, $\text{NH}4\text{NO}3$. Calculate the molecular mass of Ammonium Nitrate and determine the percentage of Nitrogen present in it. (Given atomic masses: $\text{N} = 14\text{ u}$, $\text{H} = 1\text{ u}$, $\text{O} = 16\text{ u}$) Calculate the total mass of Ammonium Nitrate (in $\text{kg}$) required for the entire $3\text{-hectare}$ field. Part B (Theoretical): Compare Manure and Fertilizers on the basis of source, nutrient concentration, effect on soil structure, and long-term environmental impact.

Part A (Numerical Solution)

1. Calculation of Molecular Mass and Nitrogen Percentage:

• Chemical formula of Ammonium Nitrate = $\text{NH}_4\text{NO}_3$

• Molecular Mass of $\text{NH}_4\text{NO}_3$: $$\text{Molecular Mass} = (2 \times \text{Mass of N}) + (4 \times \text{Mass of H}) + (3 \times \text{Mass of O})$$ $$\text{Molecular Mass} = (2 \times 14) + (4 \times 1) + (3 \times 16)$$ $$\text{Molecular Mass} = 28 + 4 + 48 = 80\text{ u}$$

• Mass of Nitrogen in 1 mole of $\text{NH}_4\text{NO}_3$ = $28\text{ u}$

• Percentage of Nitrogen in Ammonium Nitrate: $$\text{Percentage of N} = \left( \frac{\text{Total Mass of N}}{\text{Molecular Mass of } \text{NH}_4\text{NO}_3} \right) \times 100$$ $$\text{Percentage of N} = \left( \frac{28}{80} \right) \times 100 = 35%$$

2. Calculation of Total Fertilizer Required:

• Nitrogen required per hectare = $100\text{ kg}$

• Total area of the field = $3\text{ hectares}$

• Total Nitrogen required for 3 hectares = $3 \times 100\text{ kg} = 300\text{ kg}$

• Since Ammonium Nitrate contains $35%$ Nitrogen by mass: $$\text{Mass of Fertilizer Required} = \frac{\text{Total Nitrogen Required}}{\text{Percentage of N}} \times 100$$ $$\text{Mass of Fertilizer Required} = \frac{300}{35} \times 100 = 857.14\text{ kg}$$

Answer: Total mass of Ammonium Nitrate required = $857.14\text{ kg}$.

Part B (Theoretical Answer)