Asample of aluminum chloride (alcl3) has a mass of 37.2 g. a. how many aluminum ions are present? b. how many chloride ions are present? c. what is the mass in grams of one formula unit (molecule) of aluminum chloride?

C)
      The mass in grams of one formula unit of Aluminum Chloride is 133.34 g/mol.

A)
      Aluminium Ions in 37.2 g of AlCl₃:

First convert mass into moles,
As,
                                         Moles  =  Mass / M.mass
Putting Values,
                                         Moles  =  37.2 g ÷ 133.34 g/mol

                                         Moles  =  0.278 moles

As one mole contain 6.022 × 10²³ Al³⁺ Ions, so 0.278 moles will contain,

                                            = 6.022 × 10²³ × 0.278 mol

                                            =  1.67 × 10²³ Al³⁺ Ions 
B)
    As one mole of AlCl₃ contain (3 × 6.022 × 10²³ = 1.80 × 10²⁴) Ions, so 0.278 mole will contain,

                                            = 1.80 × 10²⁴ × 0.278

                                            = 5.022 × 10²³ Cl⁻ Ions

The correct answer is d