no of mole of AgNo3 added = 31.32/1000*0.1001 = 0.00313
mole
AgNO3(aq) + Cl^-(aq) ----> AgCl(s) + NO3^-
(aq)
no of mole of Cl^- present in solution =Â Â no
of mole of AgNo3 added = 0.00313 mole
no of mole of AlCl3 present = 0.00313/3 = 0.00104
mole
mass of AlCl3 present  = n*Mwt
                                 Â
= 0.00104*133.34
                                 Â
= 0.14 g
% of AlCl3 present in sample = weight of AlCl3 / sample
weight * 100
                                                      Â
= 0.14/0.9628*100
                                                   Â
= 14.54%
