The number of moles of Al₂(SO₄)₃ is 1.2 mol
Explanation:
Given:
Volume of the solution, V = 501 mL
V = 0.501L
Molarity of the solution, M = 2.37M
Moles, n = ?
We know:
Molarity = [tex]\frac{moles of solute}{volumeof solution}[/tex]
On substituting the values we get:
[tex]2.37 = \frac{n}{0.501} \\\\n = 1.2 mol[/tex]
Therefore, the number of moles of Al₂(SO₄)₃ is 1.2 mol
