# How many atoms are in 931.6 g KMnO_4?

Molar quantity of $K M n {O}_{4}$ $=$ $\frac{931.6 \cdot g}{158.034 \cdot g \cdot m o {l}^{-} 1}$ $=$ $5.89 \cdot m o l$.
There are thus $5.89$ $m o l$ $\times$ ${N}_{A}$ $\times$ $\text{6 atoms}$ $=$ $35.37 \cdot m o l$ $\times$ ${N}_{A} \text{ atoms}$ $=$ ??
${N}_{A} = \text{Avogadro's number} , 6.022 \times {10}^{23} \cdot m o {l}^{-} 1$