# What molar quantity is represented by 3.011xx10^23*"magnesium atoms"?

Dec 29, 2017

If I have a half a dozen magnesium atoms, how many atoms would that be?

#### Explanation:

And this question is PRECISELY the same as the one you asked EXCEPT that we use $\text{the dozen}$ as our metric instead of the $\text{Avocado number}$, i.e. ${N}_{A} = 6.022 \times {10}^{23} \cdot m o {l}^{-} 1$

And so we take the product...

$6.022 \times {10}^{23} \cdot m o {l}^{-} 1 \times 0.5 \cdot m o l \equiv 3.011 \times {10}^{23} \cdot \text{magnesium atoms}$. You know the mass of this quantity....

Dec 29, 2017

$3.01 \times {10}^{23}$ $\text{atoms}$.

#### Explanation:

$6.02 \times {10}^{23}$ (Avogadro's constant) indicates the number of atoms in a mole of any element, so basically:

$\text{1 mole of Mg " -> " } 6.02 \times {10}^{23}$ $\text{atoms}$
$\text{0.5 moles of Mg " -> " " "? atoms}$

So

$\text{0.5 moles" xx ( 6.02xx10^23 "atoms")/"1 mole}$

which equals to half of Avogadro's constant.