Well, let's see if I can calculate that for you without clowning around too much. To determine the amount of 5 M MgCl2 you need to add, we can use the equation:
(C1)(V1) + (C2)(V2) = (Cf)(Vf)
Where C1 and V1 are the initial concentration and volume, C2 and V2 are the concentration and volume of the 5 M MgCl2, Cf is the final concentration, and Vf is the final volume.
Given that C1 = 0.5 M, V1 = 25 ml, Cf = 1.75 M, and Vf = 1500 ml, we can solve for V2, which represents the volume of the 5 M MgCl2 to be added.
(0.5 M)(25 ml) + (5 M)(V2) = (1.75 M)(1500 ml)
12.5 + (5 M)(V2) = 2625
5 M(V2) = 2612.5
V2 ≈ 522.5 ml
So, you would need to add approximately 522.5 ml of the 5 M MgCl2 solution to achieve a final concentration of 1.75 M.
As for the amount of water, to find that out we can subtract the combined volume of the solutions (25 ml + 522.5 ml) from the final volume (1500 ml).
Amount of water = 1500 ml - (25 ml + 522.5 ml)
Amount of water ≈ 952.5 ml
Therefore, you would need to add roughly 522.5 ml of the 5 M MgCl2 solution and approximately 952.5 ml of water.