Answer to Question #116060 in General Chemistry for AL

Question

If 80.0 mL of a 0.500 M sodium carbonate, Na2CO3, solution, is diluted to 140.0 mL of solution, what would the final molarity be?

Answer 1

C1(Na2CO3)=n/V1 => n1=C1*V1=0,5*0,08=0,04

C2(Na2CO3)=n/V2=0,04/0,140=0,286