The answers to your questions are as follows
A) The number of aluminium ions present = 1.62 * 10²³ ions
B) The number of chloride ions present = 9.86 * 10²³ ions
C) The mass of one unit of aluminium chloride = 133.34 grams
<u>Given data :</u>
mass of aluminium chloride = 37.2 g
molar mass of aluminium chloride = 133.34 g/mol
note : I mole of a molecule has 6 * 10²³ molecules
Number of moles = mass / molar mass = 0.27 moles
<h3>Determine the number of aluminum ions and chloride ions present</h3>
A) aluminium ions present
moles of AlCl₃ * 6 * 10²³
= 0.27 * 6 * 10²³ = 1.62 * 10²³ ions
B) Chloride ions present
moles of AlCl₃ * 6 * 10²³ * 3
= 0.27 * 6 * 10²³ * 3
= 4.86 * 10²³ ions
C) The mass of one formula unit of aluminium chloride = 133.34 grams
Hence we can conclude that the answers to your question are as listed above.
Learn more about aluminium chloride : brainly.com/question/12849464