A solution contains 12.6 ppm of dissolved Ca(NO3)2. Calculate the concentration of NO3  in moles per litre. (4)

Md Ariful Islam

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Solution:

Each mol of Ca(NO3)2 (MM 164.088) contains 2 mol NO3- (MM 62.005),

so the fraction of mass that is nitrate (NO3-) is:

(2 mol NO3- / 1 mol Ca(NO3)2) × (62.005 g NO3- g / 164.088 g Ca(NO3)2) = 0.75575 g NO3-/ g Ca(NO3)2

If the dissolved Ca(NO3)2 has a concentration of 12.6 ppm, the concentration of dissolved NO3- is:

(0.75575) × (12.6 ppm) = 9.5225 ppm = 9.52 ppm.

1 ppm = approximately 1 mg/L

So, 9.52 ppm = 9.52 mg/L = 0.00952 g/L

The molar mass of NO3- is 62.005 g/mol.

Therefore,

(0.00952 g NO3-/ 1L) × (1 mol NO3- / 62.005 g NO3-) = 0.0001535 mol/L = 1.54×10-4 mol/L

Answer: The concentration of NO3- is 1.54×10-4 mol/L

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