Final answer:
To find the number of moles of aluminum ions in a 430mL of 1.5M Aluminum Fluoride solution, we simply multiply the volume (converted to liters) by the molarity, resulting in 0.645 moles of aluminum ions.
Step-by-step explanation:
To calculate the number of moles of aluminum ions present in a 430mL solution with a concentration of 1.5M Aluminum Fluoride (AlF3), we first need to know the stoichiometry of the compound. Aluminum Fluoride dissociates into one aluminum ion (Al3+) and three fluoride ions (F-) in solution. Therefore, the molarity of aluminum ions is the same as the molarity of the Aluminum Fluoride solution.
Next, we calculate the moles of aluminum ions using the molarity formula:
Moles of Al3+ = Molarity (M) × Volume (L)
Here, we convert the volume from milliliters to liters:
430 mL = 0.430 L
Now, we plug the values into the formula:
Moles of Al3+ = 1.5 mol/L × 0.430 L
The calculation yields:
Moles of Al3+ = 0.645 moles
Thus, there are 0.645 moles of aluminum ions in the given solution of Aluminum Fluoride.