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

$c=\frac{n}{V}$
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.