How many moles of chloride ions are present in 0.300L of a 0.150 m solution of AlCl3
To find the number of moles of chloride ions in the solution of AlCl3, we first need to determine the moles of AlCl3 present in the solution.
First, we calculate the moles of AlCl3:
Molarity = moles of solute / liters of solution
0.150 M = moles of AlCl3 / 0.300 L
moles of AlCl3 = 0.150 M x 0.300 L = 0.045 moles
In 1 mole of AlCl3, there are 3 moles of chloride ions (since AlCl3 dissociates into 1 Al3+ ion and 3 Cl- ions).
Therefore, the number of moles of chloride ions in the solution is:
0.045 moles of AlCl3 x 3 = 0.135 moles of Cl- ions
So, there are 0.135 moles of chloride ions present in 0.300 L of a 0.150 M solution of AlCl3.