How many moles are in a 501mL of a 2.37 M solution of Al2(SO4)3?

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 =

On substituting the values we get:

Therefore, the number of moles of Al₂(SO₄)₃ is 1.2 mol

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