Answer to Question #204300 in General Chemistry for Madelyn Kanehl

Question #204300

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


1
Expert's answer
2021-06-08T11:17:02-0400

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.



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