# How many grams are in 5.7 mol potassium permanganate?

There are about $900 \cdot g$ in a $5.7$ molar quantity of $K M n {O}_{4}$.
The required mass is simply the product, $\text{moles "xx" molar mass}$, i.e. 5.7*cancel(mol)xx158.034*g*cancel(mol^-1)=??*g.