A solution of common salt when added to silver nitrate solution yields a precipitate of silver chloride(0.28 g). Find the mass of sodium chloride in the solution.
0.114 g
NaCl + AgNO3 → AgCl + NaNO3
(23+35.5) (108+35.5) (23+14+48)
58.5 143.5 85
Since, 143.5 g of silver chloride is formed from 58.5 g of NaCl,
0.28 g of silver chloride is formed = 58.5×0.28143.5
= 16.38143.5
= 0.114 g of NaCl