Question

What weight of silver nitrate is required to prepare 300mL of 0.1 ppm silver (Ag+)

Answer 1

Dissociation of silver nitrate in solution:
AgNO3 ↔ Ag+ + NO3-
This means that the mol number of silver nitrate is the same for silver
n(AgNO3)=n(Ag+)
c(AgNO3)=n(AgNO3)V
n(Ag+)=n(AgNO3)=c(AgNO3)∙V
Ppm to mol/L:
ppm of silver nitrate is 5.9·10-7 mol/L
n(AgNO3)=5.9∙10-7molL∙0.3 L=1.77∙10-7mol
The mass of silver nitrate is:
m(AgNO3)=n(AgNO3)∙M(AgNO3)
m(AgNO3)=1.77∙10-7mol ∙169.87gmol=3.0∙10-5g