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Question

5.85 g of NaCl was treated with concentrated H2SO4 and the gas evolved was passed into a solution of silver nitrate. The white precipitate obtained was filtered, dried and weighed. Assuming complete reaction, how many grams of precipitate was obtained?
[Atomic mass of Ag=108 u, Na=23 u and Cl=35.5 u]

A
10.8g
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B
14.45g
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C
35.5g
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D
3.65g
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Solution

The correct option is B 14.45g
Molecular weight of Nacl=58.5 g Moles in 5.85g=5.8558.5=0.1 moles
Nacl + conc. H2SO4NaHSO4+HCl
0.1 moles
According to the above reaction,
0.1 moles of Nacl will form
0.1 moles of HCl gas.
when HCl gas pass-through
silver nitrate solution,
Following reaction will
occur.
AgNO3 + HCL Agcl + HNO 3
In the above reaction 0.1 mol
AgCl will be formed as
white pet.
Mass of 0.1 moles agcl
=0.1×(108+35.5)=0.1×143.5=14.35 g
Hence, 143.5 grams of white
ppt. will be formed.

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