An impure sample of sodium chloride which weighed 1.2 gram gave on treatment with excess of silver nitrate solution 2.4 gram of silver chloride as the precipitate. Calculate the percentage purity of the sample.

Explanation

`underset(?)underset(58.5)(NaCl + AgNO_(3)) to underset(2.40 g)(AgCl + NaNO_(3))`
143.5 g of Agcl is obtained from 58.5 g of pure NaCl 2.40 g of AgCl is obtained from pure NaCl
`=58.5/143.5 xx 2.40 = 0.978 g`
Now, 1.20 g of NaCl contain pure, NaCl = 0.978 g
100 g of NaCl contain pure NaCl `=(0.978)/1.20 xx 100 = 81.5`g
Percentage purity of sample = 81.5%