To calculate the volume of water that must be added to the mixture, we need to use the formula for dilution:

C1V1 = C2V2

Where:
C1 = initial concentration of the solution
V1 = initial volume of the solution
C2 = final concentration of the solution
V2 = final volume of the solution

In this case, we have a mixture of two different concentrations of hydrochloric acid (HCl). Let's break down the problem into smaller steps:

Step 1: Calculate the moles of HCl in the initial mixture
To find the moles of HCl, we need to use the formula:

moles = concentration (M) x volume (L)

For the 6M HCl:
moles1 = 6 M x 0.25 L = 1.5 moles

For the 2M HCl:
moles2 = 2 M x 0.75 L = 1.5 moles

Step 2: Calculate the final volume of the solution
We can rearrange the dilution formula to solve for V2:

V2 = (C1V1) / C2

Given:
C1 = 2.5 M (final concentration)
V1 = 1 L (total volume of the initial mixture)
C2 = 6 M (concentration of the 6M HCl)

V2 = (2.5 M x 1 L) / 6 M = 0.4167 L

Step 3: Calculate the volume of water to be added
Since the final volume of the solution is the sum of the initial volume and the volume of water added, we can subtract the initial volume from the final volume to find the volume of water required:

Volume of water = V2 - V1 = 0.4167 L - 1 L = -0.5833 L

The negative value indicates that we need to add 0.5833 L or 583.33 ml of water to the mixture.

Therefore, the volume of water that must be added to the mixture of 250 ml of 6M HCl and 750 ml of 2M HCl to obtain a 2.5 M solution is approximately 583.33 ml.

Use n1v1=n2v2
n=normality
v=volume