Question

How many atoms are in 3456 grams of (NH4)3PO3?

Answer 1

Firstly we will convert this mass to moles then form moles to atoms:

`\begin{gathered} moles=\frac{mass}{molar\text{ }mass} \\ _n(NH_4)_3PO_3=\frac{3456\text{ }g}{149.09\text{ }gmol^(-1)} \\ _n(NH_4)_3PO_3=23.18mol \end{gathered}`

We will now convert moles to atoms:

`\begin{gathered} Avogadro^(\prime)s\text{ }Number:6.02*10^(23)atoms \\ 1mol(NH_4)_3PO_3=6.02*10^(23)atoms \\ 23.18\text{ }mol(NH_4)_3PO_3=x\text{ }atoms(NH_4)_3PO_3 \\ 23.18\text{ }mol(NH_4)_3PO_3*(6.02*10^(23)atoms)/(1mol(NH_4)_3PO_3)=1.40*10^(25)atoms \end{gathered}`

Answer: 1.40x10^25 atoms (NH4)3PO3