# What molar quantity is associated with 19.32*g mass of magnesium metal?

Sep 6, 2017

Well, we know that $1 \cdot m o l$ of magnesium has a mass of $24.31 \cdot g$...so..............

#### Explanation:

.......so we take the quotient....

$\text{Moles of Mg} = \frac{19.32 \cdot g}{24.31 \cdot g \cdot m o {l}^{-} 1} \cong 0.80 \cdot m o l$....

....which of course is dimensionally consistent...........in that we get $\frac{\cancel{g}}{\cancel{g} \cdot m o {l}^{-} 1} = \frac{1}{\frac{1}{m o l}} = m o l$ as required.........i.e. $\frac{1}{\frac{1}{x}} = x$

What is the physical significance of this quantity? It tells us that there are $0.8 \cdot m o l \times {N}_{A} = 0.80 \cdot m o l \times 6.022 \times {10}^{23} \cdot m o {l}^{-} 1$ individual particles of magnesium in the sample......

By this reasoning, there are ${N}_{A}$ individual particles of magnesium in the sample whose mass is $24.31 \cdot g$. How did I know these numbers?