Thank you for visiting If 97 7 mL of silver nitrate solution reacts with excess potassium chloride solution to yield 0 326 g of precipitate what is the molarity. This page is designed to guide you through key points and clear explanations related to the topic at hand. We aim to make your learning experience smooth, insightful, and informative. Dive in and discover the answers you're looking for!
Answer :
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⺠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⺠in the original solution, divide the moles of Ag⺠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⺠= 0.00227 moles / 0.0977 liters, which equals approximately 0.0232 M.
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Rewritten by : Jeany
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