Commercial NaOH weighing 30g has some NaCl in it. The mixture on dissolving in water and treatment with excess AgNO3 soln. Formed a ppt weighing 14.3 g. What is the % of NaCl in sample of NaOH. [Relative mol mass of NaCl=58, AgCl=143]

Dear Student
Let the mass of NaCl in the commercial sample of NaOH be x g. 

​The NaCl present in the aqueous solution of the mixture would react with AgNO₃ solution to give white precipitate of AgI.
Molar mass of NaCl = 23 + 35.5 =​58.5g/mol

Molar mass of AgI = 107.8 + 35.5 =143.3 g/mol
NaCl (aq) + AgNO₃ (aq)​$\to$AgCl (s) + NaNO₃ (aq) 
58.5 g/mol                         143.3 g/mol

Thus according to this reaction, 
58.5 g NaCl give precipitate = 143.3 g 

Therefore x g NaCl will give precipitate = $\frac{143.3}{58.5}\times \mathrm{x} \mathrm{g}$
                                                           = 2.449  x g

​Since mass of precipitate obtained = 14.3 g 
Thus 2.449  x = 14.3
$\Rightarrow$ x = $\frac{14.3}{2.449}$
$\Rightarrow$ x = 5.839 g

Therefore amount of NaCl present in 30 g of commercial sample of NaOH = 5.839 g
​Percentage of NaCl in sample =$\frac{5.839}{30}\times 100$ = 19.46 %
Regards