In order to dilute 1.0L of a 6.00M solution of sodium hydroxide to 0.500M concentration, how much water (in mL) must you add?

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.

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