How many g of AlCl3 are in 63.1 mL of 4.20 M solution?

Question

How many g of AlCl₃ are in 63.1 mL of 4.20 M solution?

Answer 1

Molarity = Moles of solute÷Liters of solution.

Determine the moles of AlCl3 and change that into grams

4.20M = moles of AlCl3 / .0631 liter

Moles of AlCl3 = 4.20*.0631 = 0.26502 moles

Change the mole of AlCl3 into grams .26502 mole×(133.3g AlCl3/1 mole of AlCl3) = 32.3 grams of AlCl3