Answered Essay: A chemist samples a river's water to measure the amount of fertilizer runoff from the area

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A chemist samples a river’s water to measure the amount of fertilizer runoff from the area farms. In a 0.0945 L sample, the chemist measures 363 mu g of nitrate. Express the concentration of nitrate in parts per million (ppm) and parts per billion (ppb). Assume the density of the river water sample is 1.00 g/mL. 1912.455 ppm 1912455 ppb

Expert Answer

Answer

volume of the river water = 0.0945 L

mass of nitrate = 363 micro g

concentration of the nitrate = 363 * 10^-6 g/0.0945 L

Concentration of the nitrate = 3.4304 * 10^-5 g/L = 3.4304 * 10^-5 g/10^3 mL

from density of the river water 1g = 1mL

concentration of nitrate = 3.4304 * 10^-5 g/1000 g of water

concentration of Nitrate = 3.4304 * 10^-2 g/10^6 g of water

1 g of solute present in 100000 g of solution is called ppm

concentration of nitrate = 3.4304 * 10^-2 ppm

1 g of solute present in 100000000 g of solution is called ppb

concentration of nitrate = 34.304 * 10^-3 ppm

= 34.304 ppb

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