Question

A 15 ml sample of 0.20 m mgcl2 is added to 45 ml of 0.4 m alcl3 . what is the molarity of cl- ions in final solution

Answer 1

No of moles of Cl- from MgCl2 = 2*0.2*0.015 = 0.006

No of moles of Cl-from AlCl3 = 3*0.045*0.4 = 0.054

Total no of moles of Cl- in the solution = 0.06

Total volume of solution = 60mL

Thus Molarity = 0.06/0.060 = 1 mol/L

Answer 2

calculating the moles of Cl- in each of the two original solutions:

for MgCl2
0.015 L X 0.2 mol/L MgCl2 X (2 mol Cl- / 1 mol MgCl2) = 6X10^-3 mol Cl-

for AlCl3
0.045 L X 0.4 mol/L AlCl3 X (3 mol Cl- / 1 mol AlCl3) = 5.4X10^-2 mol Cl-

Adding those together gives a total of 0.060 moles Cl-

Molarity Cl- = 0.060 mol Cl- / 0.060 L = 1 M Cl-