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Sodium iodate is treated with calculated...

Sodium iodate is treated with calculated amount of sodium bisulphite to prepare one mole of iodine. How many moles of sodium disulphite are required to prepare one mole of iodine form sodium iodate ?

Text Solution

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The correct Answer is:
5
`2NaIO_(3)+underset(5mol)5NaHSO_(3)rarr 3NaHSO_(4) +2Na_(2)SO_(4)+H_(2)O+I_(2) underset(1mol)`
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Knowledge Check

  • The weight of sodium carbonate required to prepare 500 ml of a semi-normal solution is

    A
    13.25 g
    B
    26.5 g
    C
    53 g
    D
    6.125 g

  • 1 mole of sodium chloride contains

    A
    `6.022 xx 10^(22)` atoms
    B
    `6.022 xx 10^(23)` atoms
    C
    `6.022 xx 10^(24)` atoms
    D
    None of these

  • The weight of sodium bromate required to prepare of solution for cell reaction,

    A
    1.56gm
    B
    1.45gm
    C
    1.23gm
    D
    1.32gm

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