    Question

# A solution of 0.2 g of a compound containing Cu2+ and C2O2−4 ions on titration with 0.02 M KMnO4 in the presence of H2SO4 consumes 22.6 mL of the oxidant. The resultant solution is neutralized with Na2CO3, acidified with dilute acetic acid and treated with excess of KI. The liberated iodine requires 11.3 mL of 0.05 N Na2SO4 solution for complete reduction . Find out the molar ratio of Cu2+ to C2O2−4 in the compound.

A
1 : 2
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B
2 : 1
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C
1 : 4
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D
4 : 1
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Solution

## The correct option is A 1 : 2The mixture of Cu2+ and C2O2−4 are reacting seperately first with KMnO4 solution and then with KI to liberate I2. It can be seen that Cu2+ cannot be oxidised and C2O2−4 cannot be reduced. So, reactions involved are: 2MnO−4+5C2O2−4+16H+→2Mn2++10CO2+8H2O 2Cu2++4I−→ Cu2I2+I2 I2+2S2O2−3→ 2I−+S4O2−6 Equivalents of KMnO4 solution, = 0.02×5×22.61000=2.26×10−3 ∴ moles of C2O2−4, = 2.26×10−32=1.13×10−3 Equivalents of Na2S2O3 solution = 0.05×11.31000=5.65×10−4 ∴ moles of Cu2+ = 5.65×10−41=5.65×10−4 ∴ molar ratio of Cu2+ to C2O2−4 = 5.65×10−41.13×10−3=12  Suggest Corrections  0      Explore more