5.82 g of a silver coin were dissolved in strong nitric acid and excess of NaCI solution was added. The silver chloride precipitated was dried and weighed 7.20 g. Calculate the percentage of silver in the coin

5.82 g of a silver coin were dissolved in strong nitric acid and exces

1.	5.82 g of a silver coin were dissolved in strong nitric acid and excess of NaCI solution was added. The silver chloride precipitated was dried and weighed 7.20 g. Calculate the percentage of silver in the coin
Answer» Solution :The corresponding CHEMICAL equations are: `underset("one mole" 107.9 g) (2HNO_(3)) to underset("one mole")(AgNO_(3)) + NO_(2) + H_(2)O` `underset("One mole")(AgNO_(3)) + NaCl to underset("one mole" 143.35 g)(AGCL) DARR + NaNO_(3)` From the equations, it is clear that one mole (107.9 g) of Ag gives one mole (143. 35 g) of AgCI. `therefore` The mass of Ag which gives 7.20 g of AgCI `=107.9/143.35 XX 7.20 = 5.42 g` `therefore` Percentage of Ag in the given coin `=5.42/5.82 xx 100 = 93.1`