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MHT CET Chemistry question

ChemistrySome Basic Concepts of ChemistryMHT CET

g of was dissolved in and the solution was treated with excess of NaCl when 2.87 g of was precipitated. The value of is

Answer: (B)

Solution

n(AgCl) = 2.87/143.5 = 0.02 mol = n(Ag), so x = 0.02 × 108 = 2.16 g; 1.08 g corresponds to 0.01

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Question ID VK-041258