Answer to Question #225159 in Chemistry for Junior

Question

A solution contains 12.6 ppm of dissolved Ca(NO3)2. Calculate the concentration 

of NO3



in moles per litre. (4)

Answer 1

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