Question

how many grams of AgCl will precipitate if 5.8g of 0.1M NaCl is added to the saturated soln of AgCl

Answer 1

Solution of AgCl is already saturated, so no extra AgCl could precipitate. I guess that saturated solution of AgNO₃ was meant. Also AgCl can't react with NaCl, so addition of NaCl to AgCl solution changes only Cl^- concentration and, due to ionic product, can only dissolve some AgCl, but not precipitate. So I assume that AgNO₃ solution was meant to be in the task.

Considering that density of NaCl solution is 1 g/cm³, then 5.8 g is a weight of 5.8 cm³ of solution.

n = C*V = 5.8/1000*0.1 = 0.00058 mol

NaCl + AgNO₃ = AgCl + NaNO₃

n(AgCl) = n(NaCl) = 0.00058 mol

m(AgCl) = 0.00058*(108+35.5) = 0.08323 g