Question

200 ml of 3M AgNO₃ and 300 ml of 2M AgNO₃ are added to a solution containing 200 ml of 10M NaCl. The amount of NaCl left unprecipitated is

Answer 1

By dilution formula

3*200+2*300=M*500 

M=1200/500 

M=22/5 

AgNO₃+NaCl ----> AgCl +NaNO₃ 

no. of moles of AgNO₃=12/5*500/1000 = 1.2mole 

no. of moles of NaCl =10*200/1000 =2 mole here the given amount of AgNO₃ is totally consumed ,so it is the limiting reagent. 

Amount of NaCl left=0.8* 51.5 

=46.8g