x g of Ag was dissolved in HNO3 and the solution was treated with excess of NaCl, when 2.87g of AgCl was precipitated. The value of x is

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1.08 g

2.16 g

2.70 g

I .62 g

answer is B.

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Detailed Solution

2Ag+2HNO3⟶2AgNO3+H22AgNO3+2NaCl⟶2AgCl+2NaNO3AgCl143.5 g≡AgNO3170 g≡Ag108 g∵143.5 gAgCl is obtained from x=108 g∴2.87 gAgCl is obtained from x=108×2.87143.5=2.16 g

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