# How many aluminum-based electrons are present in a salt where there is an 81*g mass of aluminum ions....?

How many aluminum-based electrons in a salt where there is an $81 \cdot g$ mass of aluminum ions....?
Well, we take the quotient, $\frac{81 \cdot g}{27 \cdot g \cdot m o {l}^{-} 1} = 3 \cdot m o l$.
And thus there are $3 \cdot m o l$ of aluminum ions, and since there are 10 electrons per $A {l}^{3 +}$ ions (for Al, Z=13), there are $30 \cdot m o l$ of aluminum electrons.