# How many molecules are in 72.00 g of C_6H_12O_6?

$0.400 \times {N}_{A}$, where ${N}_{A}$ $=$ $\text{Avogadro's number}$
Given that there are $72.00$ $g$, this represents a molar quantity of $\frac{72.00 \cdot \cancel{g}}{180.16 \cdot \cancel{g} \cdot m o {l}^{-} 1}$ $=$ $0.400$ $m o l$.
We simply multiply this molar quantity by ${N}_{A} \text{ Avogadro's number}$:
$0.400 \cdot m o l \times 6.022 \times {10}^{23} \cdot m o {l}^{-} 1$ $=$ ??