# How many molecules are there in 65 g of silver nitrate AgNO_3?

Apr 4, 2016

The molar quantity of silver nitrate is $\frac{65.9 \cdot g}{169.87 \cdot g \cdot m o {l}^{-} 1}$
There are thus $\frac{65.9 \cdot g}{169.87 \cdot g \cdot m o {l}^{-} 1}$ $\times$ ${N}_{A}$ individual formula units of $A g N {O}_{3}$, where ${N}_{A} = \text{Avogadro's number} ,$ $6.022140857 \times {10}^{23} \cdot m o {l}^{-} 1$.
Thus there are $\frac{65.9 \cdot g}{169.87 \cdot g \cdot m o {l}^{-} 1}$ $\times$ ${N}_{A}$ silver atoms, the same number of nitrogen atoms. and $\frac{65.9 \cdot g}{169.87 \cdot g \cdot m o {l}^{-} 1}$ $\times$ $3 \times {N}_{A}$ oxygen atoms, because of the consitutional makeup of $A g N {O}_{3}$.