# Question #47cde

Jul 25, 2017

Well there are $1.5 \cdot m o l \times {N}_{A} \ldots \ldots . .$ carbon atoms.......

#### Explanation:

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

So we gots $1.5 \cdot \cancel{m o l} \times 6.022 \times {10}^{23} \cdot \cancel{m o {l}^{-} 1} \cong 9 \times {10}^{23} \cdot$ $\text{carbon atoms}$.

What is the mass of this quantity of carbon atoms?