no of moles of Ag+Â Â = molarity * volume in
L
                             Â
= 0.01005*0.0093Â Â = 9.3465*10-5 moles
Ag+ + Cl- -------> AgCl
since Ag+ and Cl- react in a 1:1 ratio
no of moles of Cl-Â Â = no of moles of
Ag+
no of moles of Cl-Â Â =
9.3465*10-5 moles
mass of
Cl-Â Â Â Â Â Â Â Â Â Â Â Â
= no of moles * gram atomic mass
                             Â
=9.3465*10-5 * 35.5 = 3.31*10-3 g
                                                              Â
= 3.31mg
                                                             Â
=3.31mg/0.1093L
              Â
[ total volume = 100+9.3 = 109.3ml =0.1093L]
                                      Â
= 30.28mg/L
