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Question
What is the mass of precipitate formed when 50 ml of 8.5% solution of AgNO3 is mixed with 100 ml of 1.865% potassium chloride solution?
Options
3.59 g
7 g
14 g
28 g
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Solution
3.59 g
Explanation:
\[\ce{AgNO3 + KCl -> KNO3 + AgCl}\]
50 mL of 8.5% solution contains 4.25 g of AgNO3
No. of moles of AgNO3 present in 50 mL of 8.5% AgNO3 solution
= `"Mass"/"Molar mass"`
= `4.25/170`
= 0.025 moles
Similarly, No of moles of KCl present in 100 mL of 1.865% KCl solution
= `1.865/74.5`
= 0.025 moles
So total amount of AgCl formed is 0.025 moles (based on the stoichiometry).
Amount of AgCl present in 0.025 moles of AgCl
= No. of moles × molar mass
= 0.025 × 143.5
= 3.59 g
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