Question

500 ml of 0.1 M AlCl3 is mixed with 500 ml of 0.1 M MgCl2 solution. Then molarity of Cl- in final solution is?​

Answer 1

Explanation:

MgCl2 solution. Then molarity of Cl- in final solution is

Answer 2

Answer:

0.4

Explanation:

The molarity of Cl^-1 in aluminum chloride

=3*0.2

=0.6

The molarity of Chloride ion in MgCl2

=2*0.1

=0.2

After mixing the volume changes

=(M1V1 + M2V2) / V1+V2

=500*0.6+500*0.2 / 1000

=0.4

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