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Question

# A solution of 0.2g of a compound containing Cu2+ and C2O2−4 ions on titration with 0.02M KMnO4 in the presence of H2SO4 consumes 22.6 mL of the oxidant. The resultant solution is neutralised with Na2CO3, acidified with dilute acetic acid and treated with excess KI. The liberated iodine requires 11.3 mL of 0.05 M Na2S2O3 solution for complete reduction. The mole ratio of Cu2+ and C2O2−4 in the compound will be :

A
1:2
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B
3:2
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C
2:3
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D
2:1
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Solution

## The correct option is A 1:2KMnO4 reacts only with oxalate.2MnO⊝4+5C2O2−4+16H⨁→2Mn2++10CO2+8H2OmEq of KMnO4=M×V×n factor=0.02×22.6×5mEq of oxalate =0.02×22.6×5Millimoles of oxalate =0.02×22.6×52(n factor for oxalate is 2)KI reacts only with Cu2+.2Cu2++4I⊝→Cu2I2+I2I2+2S2O2−3→2I⊝+S4O2−4mEq of thiosulphate=M×V×n-factor =0.05×11.3×1mEq of iodine=0.05×11.3×1mEq of Cu2+=0.05×11.3×1mmoles of Cu2+=0.05×11.3×1(n factor for Cu2+ is 1)mmolesofCu2+mmolesofoxalate=0.05×11.3×10.02×22.6×52=12

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