Since the concentration of the Chloride ions required is 0.100 M . From the formula of the CaCl2, it is clear that the concentration of Chloride ion produced will be double the concentration of CaCl2 ion taken.
So the concentration of the CaCl2 required = 0.100/2
= 0.05 M
Now,
let us suppose that the volume of the 3.00 M solution required = V1
Concentration of the solution to be diluted = M1= 3.00 M
Volume of the diluted solution = V2 = 250 mL
Concentration of the diluted CaCl2 solution = 0.05 M ( M2)
Now using the dilution solution ;
M1 V1. =. M2 V2
or , V1 = (M2 V2 )/M1
= (250 mL x 0.05 M )/ 3.00 M
= 4.16666667 mL
= 4.17 mL
Hence, the volume of the CaCl2 solution required = is 4.17 mL.
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