Question

A soluble iodide was dissolved in water. then, an excess of silver nitrate, agno3, was added to precipitate all of the iodide ion as silver iodide, agi. if 1.900 g of the soluble iodide gave 0.7158 g of silver iodide, how many grams of iodine are in the sample of soluble iodide?

Answer 1

Molecular weight of Ag = 107.87 Molecular weight of I = 126.9 Molecular weight of AgI = 107.87 + 126.9 = 234.77 Percentage of I in AgI is = (126.9/234.77) x 100 = 54.05% So 0.7158g of AgI contains .5405x0.7158g of soluble Iodide, which is 0.3869 grams. The answer is 0.3869 grams of iodide.