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Question

500 ml of 0.1 M KCl, 200 ml of 0.01 M NaNO3 and 500 ml of 0.1 M AgNO3 was mixed. What is the total number of moles of ions in solution?


A

[] = 0.04, [] = 0.04, [] = 0.002, [] = 0.04 []=0.042

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B

[] = 0.04, [] = 0.00166, [] = 0.04166

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C

[] = 0.04, [] = 0.00166, [] = 0.04166

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D

[] = 0.05, [] = 0.0025, [] = 0.0525

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Solution

The correct option is B

[] = 0.04, [] = 0.00166, [] = 0.04166


In this question, AgNO3 will react with KCl and AgCl forms a precipitate.
We know that,
Molarity = Number of moles of soluteNumber of litres of solution
No. of moles of KCl =0.1×5001000=0.05
No. of moles of NaNO3=2001000×0.1=0.002

No. of moles of nitrate ion from AgNO3=NO3=5001000×0.01=0.05
Hence, number of moles of AgCl precipitated = 0.05

Total nitrate ions: 0.05 + 0.002 = 0.052
Hence, in solution, only K+ ions and NO3 will be present apart from Na+.

Total moles = 0.05 + 0.052 + 0.002 = 0.104 or 104 mmoles


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