# How many atoms are present in 406.4 g of lead?

May 18, 2017

Well in $207.2 \cdot g$ of lead atoms, there are $\text{Avogadro's number}$ of particles.

#### Explanation:

And $\text{Avogadro's Number} \equiv {N}_{A} = 6.022 \times {10}^{23} \cdot m o {l}^{-} 1$........

And thus in a mass of $406.4 \cdot g$ lead atoms, there are...........

$\frac{406.4 \cdot g}{207.2 \cdot g \cdot m o {l}^{-} 1} \times 6.022 \times {10}^{23} \cdot m o {l}^{-} 1$

$= 1.18 \times {10}^{24}$ $\text{individual lead atoms} .$

This use of mass and Avogadro's number to define the number of particles, of atoms, and molecules, is fundamental to the study of quantitative chemistry.