Answer:
0.15M
Explanation:
Molar Concentration = [tex]\frac{mass}{molar mass}[/tex]
Assuming 20g of the unadulterated aluminium chloride was weighed into the volumetric flask, and given molar mass of AlCl3 = 133.34g/mol
∴ Molarity = [tex]\frac{20}{133.34}[/tex]
= 0.14999 ≈ 0.15M (to 2 significant figures)
I hope this solution is clear. The same can be calculated from concentration by volume.