Question

calculate the molarity of a solution prepared by dissolving 6.80 grams of AgNO3 in enough water to make 2.50 liters of solution.

Answer 1

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)

Answer 2

`c=(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 \ (g)/(mol) \\ m=6.8 \ g \\ n=(m)/(M)=(6.8 \ g)/(170 \ (g)/(mol))=0.04 \ mol`

The volume is
`V=2.5 \ L`.

The molarity:

`c=(0.04 \ mol)/(2.5 \ L)=0.016 \ (mol)/(L)`.

The molarity is 0.016 mol/L.