Afarmer applies 1705 kg of a fertilizer that contains 10.0% nitrogen to his fields each year. Fifteen percent (15.0%) of the fertilizer washes into a river that runs through the farm.
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If the river flows at an average rate of 0.455 cubic feet per second, what is the additional concentration of nitrogen (expressed in
milligrams of nitrogen per liter) in the river water due to the farmer's fertilizer each year?
mg/L

The additional concentration of nitrogen in the river water due to the farmer's fertilizer is 0.0629 milligrams per liter

Explanation:

The mass of the fertilizer the farmer applies = 1705 kg

The percentage of nitrogen contained in the fertilizer = 10%

Therefore;

The mass of nitrogen contained in the fertilizer = 10% of 1705 kg = 10/100 × 1705 kg

The mass of nitrogen contained in the fertilizer = 10/100 × 1705 kg = 170.5 kg

The percentage of the fertilizer that washes into the river that runs through the farm = 15.0%

The mass of the fertilizer that washes into the river = 15/100 × 170.5 = 25.575 kg = 25575 g = 25575000 mg

The volume of water that flows through the farm = 0.455 cubic feet per second

The volume of water that flows through the farm in a year = 0.455 ft³ × 60 × 60 × 24 × 365 = 14348880 ft³ = 406,315.03 m³ = 406,315,034 l

The number of moles of nitrogen = Mass of nitrogen/(Molar mass of nitrogen)

Molar mass of nitrogen =14.0067 g/mol

Therefore;

The number of moles of nitrogen = 25575 kg/(14.0067 g/mol) = 1.825.912 moles

The additional concentration of the nitrogen in moles = 1.825.912 moles/(406,315,034 l) = 4.49 ×   10⁻⁶ mol/liter

The additional concentration of the nitrogen in milligrams of nitrogen per liter of water = 25575000/406,315,034 = 0.0629 milligrams of nitrogen per liter.

Therefore, the additional concentration of nitrogen in the river water due to the farmer's fertilizer = 0.0629 milligrams per liter

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