\(c=\frac{n}{V}\)
c - the molarity, n - the number of moles, V - the volume of the solution
First calculate the number of moles of AgNO₃ in the solution.
\(AgNO_3 \\ M=170 \ \frac{g}{mol} \\
m=6.8 \ g \\
n=\frac{m}{M}=\frac{6.8 \ g}{170 \ \frac{g}{mol}}=0.04 \ mol\)
The volume is \(V=2.5 \ L\).
The molarity:
\(c=\frac{0.04 \ mol}{2.5 \ L}=0.016 \ \frac{mol}{L}\).
The molarity is 0.016 mol/L.
