A 5.82 g silver coin is dissolved in nitric acid. When sodium chloride is added to the solution, all the silver is precipitated as AgCl . The AgCl precipitate weighs 7.20 g . The percentage of silver in the coin is

Question

A 5.82 g silver coin is dissolved in nitric acid. When sodium chloride is added to the solution, all the silver is precipitated as AgCl . The AgCl precipitate weighs 7.20 g . The percentage of silver in the coin is (A)  60.3 %  (B)  80 %  (C)  93.1 %  (D)  70 %

Tardigrade Admin · Accepted Answer

Silver solution in HNO3+NaCl → underset108+35.5AgCl Atomic mass of Ag = 108 g =143.5 Weight of Ag = 5.82 g ∵ 143.5 g AgCl is obtained from Ag = 108 g ∴ 7.20 g AgCl will be obtained from Ag =(108× 7.20/143.5)=5.418 g Weight of Ag taken =5.82 % purity =(5.418/5.82)×100 =93.09 %

Q. A $5.82 g$ silver coin is dissolved in nitric acid. When sodium chloride is added to the solution, all the silver is precipitated as $A g Cl$ . The $A g Cl$ precipitate weighs $7.20 g$ . The percentage of silver in the coin is

$60.3%$

$80%$

$93.1%$

$70%$

Q. A 5.82g silver coin is dissolved in nitric acid. When sodium chloride is added to the solution, all the silver is precipitated as AgCl . The AgCl precipitate weighs 7.20g . The percentage of silver in the coin is

60.3%

80%

93.1%

70%

Silver solution in HNO3​+NaCl→108+35.5AgCl​ Atomic mass of Ag=108g=143.5 Weight of Ag=5.82g ∵ 143.5gAgCl is obtained from Ag=108g ∴7.20gAgCl will be obtained from Ag =143.5108×7.20​=5.418g Weight of Ag taken =5.82 % purity =5.825.418​×100 =93.09%

Q. A $5.82 g$ silver coin is dissolved in nitric acid. When sodium chloride is added to the solution, all the silver is precipitated as $A g Cl$ . The $A g Cl$ precipitate weighs $7.20 g$ . The percentage of silver in the coin is

$60.3%$

$80%$

$93.1%$

$70%$