10-2 mol of Fe3O4 is treated with excess of KI solution in the 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)?

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50 ml

100 ml

200 ml

400 ml

answer is C.

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Detailed Solution

∴ Fe3O4+2I−→Fe2++I2∴ No. of moles of I2 produced =10−2mol Let vml0.1(M)Na2S2O3 solution is required ∴ v×10−4=2×10−2or v=200 ml

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