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A water sample contains 10 mg NO₃ - /L. What is the concentration in_____

(a) ppmm
(b) moles/L
(c) mg NO₃ - -N/L
(d) ppbm?

User My Helper
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Final answer:

The concentration of nitrate in the water sample is 10 ppm, which is equivalent to 0.0001613 mol/L, 2.26 mg N/L, and 10,000 ppb.

Step-by-step explanation:

The concentration of nitrate (NO3−) in a water sample is given as 10 mg NO3−/L. We can express this concentration in various units such as parts per million (ppm), moles per liter (mol/L), and parts per billion (ppb).

(a) Concentration in ppm

There doesn't seem to be a standard unit called 'ppmm,' so this part may be a typo or misunderstanding. Usually, we refer to ppm (parts per million) which, in aqueous solutions, is equal to mg/L. Therefore, the concentration is 10 ppm since 1 ppm = 1 mg/L.

(b) Concentration in moles/L

To convert to moles per liter, we need to know the molar mass of NO3−, which is approximately 62 g/mol. Thus, the conversion is:

  • 10 mg NO3− / L × (1 g / 1000 mg) × (1 mol / 62 g) = 0.0001613 mol/L

(c) Concentration in mg NO3−-N/L

The nitrogen atom in a nitrate ion (NO3−) contributes to part of the total mass. The molar mass of nitrogen (N) is approximately 14 g/mol. Therefore, the conversion of mg NO3−/L to mg NO3−-N/L is:

  • 10 mg NO3− / L × (14 g/mol N / 62 g/mol NO3−) = 2.26 mg N/L

(d) Concentration in ppb

Converting ppm to ppb (parts per billion), we know that 1 ppm is 1,000 times less than 1 ppb. Therefore:

  • 10 ppm × 1000 = 10,000 ppb
User Arjaan Buijk
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