Question 10: If the solubility of Barium Chloride is 37.5g per 100mL, what is the molarity of a solution prepared by mixing 315g of barium chloride with water to make 750mL of solution?

First I took the 315g BaCl2 and divided it by 207g BaCl2(Atomic Weight.) to find out how many Moles of BaCl2 there were, and I came up with 1.52Mol BaCl2. Then I tool the 750mL and divided that by 1000 to get .75L of solution. I took the 1.52Mol BaCl2 and divided that by .75L and got the molarity of 2.02 but the sheet says the correct answer is 1.81. Am I missing a step somewhere?