Gyou have 57.0 ml of a 0.400 m stock solution that must be diluted to 0.100 m. assuming the volumes are additive, how much water should you add?

added water = 171 ml

Explanation:
Assuming volumes are additive, the rule that we will use to solve this question is:
M1V1 = M2V2
where:
M1 is the initial concentration = 0.4 m
V1 is the initial volume = 57 ml
M2 is the final concentration = 0.1 m
V2 is the final volume that we want to calculate
Substitute with the given in the above equation to get V2 as follows:
M1V1 = M2V2
(0.4)(57) = (0.1)V2
22.8 = 0.1V2
V2 = 228 ml

Now, the final volume is equal to the initial volume plus the amount of added water. So, to get the amount of added water, we will subtract the initial volume from the final volume as follows:
V2 = V1 + added water
228 = 57 + added water
added water = 228 - 57 = 171 ml

Hope this helps :)

Mgo2 is the oxidation state of manganese oxide