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A test for the complete removal of Cu2+ ions from a solution of Cu2+ (aq.) is to add NH3 (aq.). A blue colour signifies the formation of complex [Cu(NH3)4]2+ that has Kf=1.1×1013 and thus confirms the presence of Cu2+ in the solution. 250 mL of 0.1 M CuSO4 (aq.) is electrolyzed by passing a current of 3.5 ampere for 1350 seconds. Then, sufficient quantity of NH3 (aq.) is added to the electrolyzed solution maintaining [NH3]=0.1 M. If  [Cu(NH3)4]2+ is detectable upto its concentration as low as 1×105. Pick the correct statement(s):


A
Amount of Cu2+deposited = 1.554 g
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B
Amount of Cu2+ left in the solution = 2.07×103
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C
Amount of Cu2+ left in the solution = 4×103
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D
Solution will show blue colour
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Solution

The correct options are
A Amount of Cu2+deposited = 1.554 g
B Amount of Cu2+ left in the solution = 2.07×103
D Solution will show blue colour
Cu2++4NH3[Cu(NH3)4]2+
               Kf=[Cu(NH3)4]2+[Cu2+][NH3]4

Blue colour will not be noticed upto [Cu(NH3)4]2+ = 1×105

Thus at this stage
Cu2+=1×1051.1×1013×(0.1)4=9.1×1015 M

Cu deposited (w)
= E.i.t.96500=63.5×3.5×13502×96500=1.5546 g

millimoles of Cu2+ present initially
= 250×0.1

Weight of Cu2+ present initially
 = 250×0.1×63.51000
 = 1.5875 g 

Weight of Cu2+ left in the solution
= 1.5875 - 1.5546 = 0.0329 g

 Cu2+ left = 0.0329×100063.5×250=2.07×103

Thus solution will show blue colour, as it will provide appreciable  Cu2+ to form complex.

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