i) 1L of 0 1 M NaCl No of moles of NaCl = 0 1 x 1 = 0 1 ii) 2L of 0 2M AgNO3 solution No of moles of AgNO3 = 0 2 x 2 = 0 4 Reaction involved : AgNO3 + NaCl → AgCl + NaNO3 Part : I Thus, 0 1moles of NaCl is added to 0 4moles of AgNO3 to produce 0 1 moles of AgCl as precipitate Part : II Weight of AgCl = 0 4 x molar mass of AgCl = 0 4 x 143 = 57 2 g Part : III Limiting reagent : NaCl as it will be consumed first due to availability of less no of moles