The volume (in ml) of  0.1 M required for complete precipitation of chloride ions present in of solution of CrH2O5Cl Cl2  as silver chloride is close to

CrH2O5Cl Cl2+2AgNO3→CrH2O5ClNO3+2AgCl1 milli mole→2 milli moles of AgNO3For 30 ml & 0.01​M⏟0.3​ milli​ moles → ?0.3×21=0.6 mill moles of AgNO3 RequiredBut mill mole=Molarity × volumein ml0.6=0.1 M ×volume in ml∴  Volume in ml =0.60.1 =6 ml