Answer :
[tex]c=\frac{n}{V}[/tex]
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.
[tex]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[/tex]
The volume is [tex]V=2.5 \ L [/tex].
The molarity:
[tex]c=\frac{0.04 \ mol}{2.5 \ L}=0.016 \ \frac{mol}{L}[/tex].
The molarity is 0.016 mol/L.
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.
[tex]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[/tex]
The volume is [tex]V=2.5 \ L [/tex].
The molarity:
[tex]c=\frac{0.04 \ mol}{2.5 \ L}=0.016 \ \frac{mol}{L}[/tex].
The molarity is 0.016 mol/L.
molar mass: 107.9gAg + 14.01gN + 3(16.00gO) = 169.91 g/mol AgNO3
6.80gAgNO3 ( mol AgNO3 )( 1 )= 0.0160M AgNO3
(169.91g AgNO3)(2.50L)
6.80gAgNO3 ( mol AgNO3 )( 1 )= 0.0160M AgNO3
(169.91g AgNO3)(2.50L)
