How many atoms are in 1.92 mol KMnO_4?

Mar 3, 2016

$1$ $m o l$ of particles specifies Avogadro's number of particles, so $\text{no. of atoms" = N_Axx1.92*molxx6" atoms } m o {l}^{-} 1 \cdot K M n {O}_{4}$.
So, ${N}_{A}$, Avogadro's number, specifies $6.022 \times {10}^{23}$ $m o {l}^{-} 1$.
You have quoted a $1.92$ $\text{mole}$ quantity. This specifies $1.92$ $\text{mole}$ $\times$ $6.022 \times {10}^{23}$ $m o {l}^{-} 1$ individual $K M n {O}_{4}$ units.
Now each $K M n {O}_{4}$ unit comprises $6$ atoms. So simply multiply the molar quantity by 6