Question

How much amount of urea would be required for supplying 5 Kg of N?

• A. 5.25
• B. 9.62
• C. 10.87
• D. 15.66

Hint:

To simplify urea calculations, multiply the target nitrogen requirement by $2.17$ (since $100 / 46 \approx 2.17$).
For example: $5\text{ kg Nitrogen} \times 2.17 = 10.85\text{ kg of Urea}$.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
Calculating fertilizer requirements involves determining the total weight of a commercial fertilizer needed to supply a specific quantity of a target plant nutrient.
Key Formula or Approach:
The formula to calculate the quantity of fertilizer is:
\[ \text{Fertilizer required (kg)} = \frac{\text{Nutrient recommended (kg)} \times 100}{\text{Nutrient percentage in fertilizer (\%)}} \]

Step 2: Detailed Explanation:
Urea is a common nitrogenous fertilizer that contains $46\%$ Nitrogen ($N$) by weight.
This means that $100\text{ kg}$ of commercial urea contains $46\text{ kg}$ of pure nitrogen.
To find the amount of urea needed to supply $5\text{ kg}$ of nitrogen, substitute the values into the formula:
\[ \text{Urea required} = \frac{5 \times 100}{46} \]
\[ \text{Urea required} = \frac{500}{46} \approx 10.87\text{ kg} \]
Therefore, $10.87\text{ kg}$ of urea is required to provide $5\text{ kg}$ of nitrogen.

Step 3: Final Answer:
The amount of urea required is $10.87\text{ kg}$, which corresponds to Option (C).