Inputs
Result
mg/l
50.0000
µg/l
50,000.00
ng/l
50,000,000
If this is nitrate as NO₃, as nitrogen
11.29mg/l as N
Divide by 4.4268
If this is nitrate as N, as nitrate ion
221.34mg/l as NO₃
Multiply by 4.4268

Method

In water at ambient conditions, 1 mg/l is taken as equal to 1 ppm. The nitrate conversion uses the ratio of molar masses: NO₃ (62.004) ÷ N (14.007) = 4.4268.