The concentration of the CaCl2 would be 0.143 M
Using the dilution equation:
M1V1 = M2V2
That is, the number of moles of solutes in a solution before dilution must be equal to the number of moles of solutes in the solution after dilution.
In this case,
M1 = 0.250, V1 = 100 mL, M2 = ? V2 = 100 + 75 = 175 mL
M2 = M1V1/V2
= 0.250 x 100/175
= 0.143 M
Thus, the diluted solution must have a molarity of 0.143 M
More on dilution can be found here: https://brainly.com/question/17145416