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Answer :
The complete balanced chemical reaction is written as:
AgNO3 + KCl ---> AgCl
+ KNO3
where AgCl is our
precipitate
So calculating for moles
of AgCl produced: MM AgCl = 143.5 g/mol
moles AgCl = 0.326 g /
(143.5 g/mol) = 2.27 x 10^-3 mol
we see that there is 1
mole of Ag per 1 mole of AgCl so:
moles Ag = 2.27 x 10^-3
mol
The molarity is simply
the ratio of number of moles over volume in Liters, therefore:
Molarity = 2.27 x 10^-3
mol / 0.0977 L
Molarity = 0.0233 M
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Final answer:
To find the molarity of silver ion in the original 97.7 ml solution that yielded 0.326 g of AgCl precipitate, calculate the moles of AgCl from its mass and then divide by the solution volume in liters, resulting in a molarity of approximately 0.0232 M.
Explanation:
If 97.7 ml of silver nitrate solution reacts with excess potassium chloride solution to yield 0.326 g of precipitate, the molarity of silver ion in the original solution can be determined following these steps:
Identify the precipitate formed which is likely AgCl (silver chloride), due to the reaction between silver nitrate (AgNO₃) and potassium chloride (KCl).
Calculate the moles of the precipitate using its mass and molar mass. The molar mass of AgCl is approximately 143.32 g/mol. Therefore, moles of AgCl = mass / molar mass = 0.326 g / 143.32 g/mol, giving approximately 0.00227 moles of AgCl.
Since the reaction of AgNO₃ with KCl produces AgCl in a 1:1 molar ratio, the moles of Ag⁺ in the original solution are the same as the moles of AgCl precipitated, which is 0.00227 moles.
Finally, to find the molarity of Ag⁺ in the original solution, divide the moles of Ag⁺ by the volume of the solution in liters. The volume of the solution is given as 97.7 mL, which is 0.0977 liters. Therefore, the molarity of Ag⁺ = 0.00227 moles / 0.0977 liters, which equals approximately 0.0232 M.
