# How many mols are in 23.7 g of KMnO_4?

$\text{Number of moles}$ $=$ $\text{Mass"/"Molar mass}$
And moles of $\text{potassium permanganate}$ $=$ $\frac{23.7 \cdot g}{158.03 \cdot g \cdot m o {l}^{-} 1}$ $\cong \frac{1}{6} \cdot m o l$.