Answer to Question #204300 in General Chemistry for Madelyn Kanehl

Question

How many grams of barium sulfate would be needed to create 0.650 liters of a 1.5 m solution

Answer 1

Molarity= moles/litres

moles= Molarity*litres

moles=1.5M*0.650L

= 0.975moles

Molecular mass*moles we get quantity in grams of barium sulfate needed.

Therefore, 233.39*0.975=227.56g of barium sulfate are needed.