# How many atoms are there in a 58.7*g mass of "ammonium nitrite"?

Sep 10, 2017

Well, there are $\frac{58.7 \cdot g}{64.06 \cdot g \cdot m o {l}^{-} 1} = 0.916 \cdot m o l$ with respect to $\text{ammonium nitrite}$.....we get approx. $42 \times {10}^{23}$ individual atoms.

#### Explanation:

And so..................

${\text{no. of atoms"="moles of ammonium nitrite"xx"7 atoms mol}}^{-} 1 \times {N}_{A}$

where ${N}_{A} = 6.022 \times {10}^{23.} \ldots \ldots .$

And so we gots 7xx6.022xx10^23*mol^-1xx0.916*mol=??