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Q.

The number of moles of NH3, that must be added to 2L of 0.80 M AgNO3 in order to reduce the concentration of Ag+ ions to 5.0x108M is (Kf for [Ag(NH3)2]+=1.0 x 108)

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answer is 4.

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Detailed Solution

Ag++2NH3[Ag(NH3)2]+

t = 01.6 mol a mol 0 mol 

t=    2×5×108mol (a 3.2) mol 1.6 mol

M  5×108M  (a3.2)/2M   1.6/2M

Given Kf=1×108

108= 0.8/5 × 10-8×[(a-3.2)/2]2

a=4

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