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# 0.1 M solution of KI reacts with excess of H2SO4 and KIO3 solutions, according to the equation 5I−+IO−3+6H+→3I2+3H2O which of the following statements is/are correct

A
200 ml of the KI solution reacts with 0.004 mole KIO3
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B
100 ml of the KI solution reacts with 0.006 mole H2SO4

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C
0.5 litre of the KI solution produced 0.005 mole of I2
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D
Equivalent weight of KIO3 is equal to (molecular weight5)
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Solution

## The correct options are A 200 ml of the KI solution reacts with 0.004 mole KIO3 D Equivalent weight of KIO3 is equal to (molecular weight5) a) 200 ml of the KI solution = 200×0.1 mmol of KI=20 mmol. So from the stoichiometry of the reaction, 200 ml of the KI solution reacts with 4 mmol of KIO3. b) 100 ml of the KI solution = 100×0.1 mmol of KI=10 mmol. So from the stoichiometry of the reaction, 100 ml of the KI solution reacts with 65×10=12 mmol of H2SO4. c) 0.5 litre of the KI solution = 500×0.1 mmol of KI=50 mmol. So from the stoichiometry of the reaction, 0.5 litre of the KI solution produced 35×50=30 mmol of I2. d) Equivalent weight of KIO3 is equal to (molecular weight of KIO3n−factor of KIO3)=molecular weight5  Suggest Corrections  3      Similar questions  Explore more