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Question

10.0 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 the liberation of I2. The solution is again filtered and boiled till all the I2 is removed. Now excess of an oxidising agent is added to the filtrate which liberates I2 again. The liberated I2 now requires 10 mL of 1.0M Na2S2O3 solution. The percentage by mass of Cu2O in the sample is :

A
27.6%
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B
28.6%
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C
28.5%
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D
none of the above
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Solution

The correct option is B 28.6%
Cu2OCu2++Cu0
The solution after dissolution of Cu2O in dil.H2SO4 contains Cu2+ ions and Cu. Cu2+ ions react with KI to give CuI2 which is converted to CuI (or Cu2I2) and I2.
Cu2++2ICuI2(Cu2I2 or CuI)+12I2
millimoles of KI taken =8.3166×103=50(Mw of KI=166)
Now, KI left unused reacts with oxidising agent to liberate I2 again.
2IOxidisingagent−−−−−−−−−I22S2O32−−−−S4O62+I
millimoles of KI left = millimoles of S2O32 used
(n=22=1)=(n=22=1) =10×1.0×1(n-factor)=10
Therefore, millimoles of KI used for Cu2+=5010=40.
Therefore, milli moles of Cu2O=20
C+1u22+2CuMolarratioofCu2:+2Cu=1:2
WeightMw×103=20 (Mw of Cu2O=143)
Weight143×103=20
WCu2O=2.86
% of Cu2O=2.86×10010.0=28.6%

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