Answer:
[tex]\large \boxed{\text{0.4200 mol}}[/tex]
Explanation:
[tex]\text{M$_{r}$ of KMnO$_{4}$} = 158.03\\\text{Moles} = \text{66.38 g } \times \dfrac{\text{1 mol}}{\text{158.03 g}} = \text{0.4200 mol}\\\\\rm \text{The sample contains $\large \boxed{\textbf{0.4200 mol}}$ of KMnO$_{4}$}[/tex]
