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Question

A 3.0 g sample of Cu2O is dissolved in dil.H2SO4 where it undergoes disproportionation quantitatively. The solution is filtered off and 8.3 g pure KI crystals are added to clear filtrate in order to precipitate CuI with evolution of I2. The solution is again filtered and boiled till all the I2 is expelled. Now excess of an oxidizing agent is added to filtrate, which liberates I2 again. The liberated I2 this time requires 10 mL of 1.0N Na2S2O3 solution. Calculate % by mass of Cu2O in sample.

A
95
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B
98
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C
92
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D
89
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Solution

The correct option is C 95
(Cu+)2Cu2++Cu
The solution after dissolution of Cu2O in dil.H2SO4 contains Cu2+ and Cu ions. Cu2+ ions react with KI to give CuI2, which is converted to CuI and I2.
Cu2++2ICuI2CuI+12I2
MIllimole of KI taken =8.3166×1000=50
Now, KI left unused reacts with oxidizing agent to liberate I2 again.
2IOxidant−−−−I22Na2S4O3−−−−−Na2S4O6+2NaI
Millimole of KI left = Millimole of Na2S2O3 used (mole ratio of I to Na2S2O3is1:1) =10×1.0=10
Therefore, millimoles of KI used for Cu2O=5010=40
Millimoles of Cu2O=20 [Cu+22Cu2+ mole ratio Cu+2:Cu2+::1:2]
or, wM×1000=20 =w×1000142=20
wCu2O=2.84
%ofCu2O=2.84×1003=94.67

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