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Q.

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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a

200 ml

b

50 ml

c

400 ml

d

100 ml

answer is C.

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Detailed Solution

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∴ Fe3O4+2I−→Fe2++I2

∴  No. of moles of I2 produced =10−2mol

 Let vml0.1(M)Na2S2O3 solution is required 

∴ v×10−4=2×10−2

or v=200 ml

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