Commercial sodium hydroxide weighing 30 g has some sodium chloride in it.
The mixture on dissolving in water and subsequent treatment with excess silver nitrate solution formed a precipitate weighing 14.3 g. What is the percentage of sodium chloride in commercial sample of sodium hydroxide? The equation for the reaction is
NaCl + AgNO3 ----> AgCl + NaNO3
[Relative molecular mass of NaCl = 58 ; AgCl = 143]
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)-----> 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 =
= 2.449 x g
Since mass of precipitate obtained = 14.3 g
Thus 2.449 x = 14.3
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 = = 19.46 %