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?
Answers
Answered by
0
Explanation:
MgCl2 solution. Then molarity of Cl- in final solution is
Answered by
1
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
Buddy please mark as Brilliant answer.. if it helped
Similar questions