Respuesta :

Another M1V1=M2V2 
V1=1.0L 
M1=6.00M 
M2=0.500M 
V2=?? 
(6.00M)(1.0L)=(0.500M)V2 
Solve for V2 you get 12L or 12000mL 
MissPhiladelphia

To solve this we use the equation, 

M1V1 = M2V2

where M1 is the concentration of the stock solution, V1 is the volume of the stock solution, M2 is the concentration of the new solution and V2 is its volume.

6.00 M ( 1.0 L ) = ( 0.500 M ) V2

V2 = 12 L

Therefore, approximately 11 L of water should be added to the 1 L of 6.00 M solution. Hope this answers the question. Have a nice day.

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