Answer to Question #258850 in Chemistry for Lay

Question

(a) How many ppm of Ca(NO3)2 are in 0.144mMCa(NO3)2?

Answer 1

Solution:

0.144 mM Ca(NO₃)₂ = 0.144 mmol Ca(NO₃)₂ / L

The molar mass of Ca(NO₃)₂ is 164.088 g/mol

Because 1 mol of Ca(NO₃)₂ is 164.088 g, by extension, 1 mmol = 164.088 mg

1 mmol = 164.088 mg

Therefore, 0.144 mmol is:

(0.144 mmol) × (164.088 mg / 1 mmol) = 23.63 mg Ca(NO₃)₂ = 23.6 mg Ca(NO₃)₂

ppm = 1 mg/L

Therefore, 23.6 mg/L = 23.6 ppm

Answer: 23.6 ppm of Ca(NO₃)₂