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Question

If one assumes the volumes are additive, what is the [Cl-] in a solution obtained by mixing 225 mL of 0.625 M KCl and 615 mL of 0.385 M MgCl2?

Answer 1

The total amount of Cl- is: 0.225*0.625+2*0.615*0.385 = 0.614(mole)
The total volume of the solution after mixing is: 0.225+0.615 = 0.840(L)
So: [Cl-] = 0.614/0.840 = 0.731(M)