10−2 moles of Fe3O4is treated with excess of KI solution in presence of dilute H2SO4, the products are Fe2+ and I2 (g). What volume of 0.1 (M) Na2S2O3 will be needed to reduce the liberated I2 (g) ?
A
50 ml
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B
100 ml
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C
200 ml
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D
400 ml
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Solution
The correct option is C 200 ml
∴Fe3O4+2I−→Fe2++I2 ∴ no.of moles of I2 produced =10−2moles Let v ml 0.1(M)Na2S2O3 solution is required ∴v×10−4=2×10−2 or v=200 ml.