Answer:
Option C. 0.34 moles
Explanation:
Data obtained from the question include:
Molarity = 0.15M
Volume = 2.25L
Mole =
Molarity = mole /Volume
Mole = Molarity x Volume
Mole = 0.15 x 2.25
Mole = 0.34mole
The number of mole of AlCl3 present in the solution is 0.34mole.
