# How many Cl^(-) ions are there in 584.6 grams of AlCl_3?

Jan 3, 2018

Well, formally there are 3 moles of chloride per mole of aluminum chloride..

#### Explanation:

And so we (i) assess the molar quantity of aluminum chloride, $= \frac{584.6 \cdot g}{133.34 \cdot g \cdot m o {l}^{-} 1} = 4.38 \cdot m o l$...

And (ii) we multiply this molar quantity by factors of THREE and Avogadro's number to get the number of chloride ligands...

$= 13.15 \cdot m o l \times \text{avocado number}$

$13.15 \cdot m o l \times 6.022 \times {10}^{23} \cdot m o {l}^{-} 1 = 7.92 \times {10}^{24} \cdot \text{chlorides}$