a liter of 8m aluminum hydroxide solution is equally divided into two half of the solution is transferred to another reagent bottle . this is diluted with water and its concentration decreases to 3.0 n while the other reagent bottle is also added with water to decrease its concentration to 6.0 n . what is the total volume of water . added to the solution ?

Question

mww · Answer

Your units are rather unclear. What does 3.0 n mean? Is this molar?

mww · Answer

In any case this type of calculation involves two things

1) Calculate the no. of moles of solute in a given volume using the formula n = cV where c is the molarity (mol/L) and V is the volume (in litres) of the solution.

Since you transferred contents into two solutions with half the volume each, V = 0.5 L

2) When you dilute such a solution, all you do is add solvent to increase the volume of the solution. So there is no change to the amount of solute in there. The same no. of moles of solute as in part 1 will be there. i.e. n(old) = n(new)

The new concentration C(new) = n/V(new) so 
V(new) = n/C(new)
The volume required to add is then V(new) - V(old) = V(new) - 0.5L

Do a similar thing with the other half and you'll find the required volume there.