The correct option is B 28.6%
Cu2O→Cu2++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++2I⊝→CuI2→(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.
2I⊝Oxidisingagent−−−−−−−−−−→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+=50−10=40.
Therefore, milli moles of Cu2O=20
⎡⎢⎣C+1u2→2+2Cu∴MolarratioofCu2:+2Cu=1:2⎤⎥⎦
WeightMw×103=20 (Mw of Cu2O=143)
Weight143×103=20
∴WCu2O=2.86
% of Cu2O=2.86×10010.0=28.6%