If your MgCl2 is %95 pure, your mgCl2 is 0,12*95/100=0,114 g
95 g/mol MgCl2 contain 71 G/mol Cl- (2*Cl-)
then 0,114 g MgCl2 contain 0,0852 g Cl-
the mol of Cl- is 0,0852/35,5= 0,0024 mol
Ag++ Cl-------AgCl
Ag: Cl= 1:1
so the mol of Ag+ is 0,0024 mol (that is the AgNO3's mol number too).
ıf my AgNO3 solution is 0,1 M and the mol number is 0,0024 mol then the volume of AgNO3 that I need is 0,0024/0,1= 0,024 L= 24 mL
and %10 excess 24+(24*10/100)= 26,4 mL
NOTE: if you add %10 ecsess of AgNO3 solution, your Cl- is precipitated completely.