Question

20 gm impure NaCl is added in excess of AgNO3. Mass of AgCl produced is 28.7 gm. Find % purity of NaCl sample (Ag = 108)

Answer 1

Answer:

NaCl + AgNO3( excess) → NaNO3 + AgCl ( ↓)

here we see that , 1 mole of NaCl precipitate in 1 mole of AgCl .

so, (23+35.5)g of NaCl precipitate in ( 108 + 35.5) g AgCl

so, 58.5 g of NaCl precipitate in 143.5 g AgCl

so, 1.2 g of NaCl precipitate in (143.5)/58.5)× 1.2 g = 2.94 g , AgCl

but in question only 2.4 g NaCl precipitate

so, impurity = 0.54 g

% impurity = (0.54)/2.94 × 100

= 18.36 %

so, % purity = 100-18.36 = 81.64 %

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