How many ml of a 14.0 m nh3 stock solution are needed to prepare 200 ml of a 4.20 m dilute nh3 solution? hints how many ml of a 14.0 m stock solution are needed to prepare 200 ml of a 4.20 m dilute solution? 0.060 ml 60.0 ml 840 ml 667 ml?

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.

14 M x V1 = 4.20 M x 200 mL

in the analysis of systems that involve fluid flow, we frequently encounter the combination of properties u and pv.