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A(2)O(X) is oxidised to AO(3)^(-) by MnO...

`A_(2)O_(X)` is oxidised to `AO_(3)^(-)` by `MnO_(4)^(-)` in acidic medium. If `1.5times10^(-3)` mole of `A_(2)O_(X)` requires 40ml of `0.03M-KMnO_(4)` solution in acidic medium. The value of `x`

Answer

Step by step text solution for A_(2)O_(X) is oxidised to AO_(3)^(-) by MnO_(4)^(-) in acidic medium. If 1.5times10^(-3) mole of A_(2)O_(X) requires 40ml of 0.03M-KMnO_(4) solution in acidic medium. The value of x by CHEMISTRY experts to help you in doubts & scoring excellent marks in Class 12 exams.

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An oxide X_(2)O_(n) is oxidised to XO_(3)^(-) by MnO_(4)^(-) in acidic medium.If 1.5times10^(-3) moles of X_(2)O_(n) requires 1.8times10^(-3) moles of MnO_(4)^(-) for complete oxidation.Assuming that oxidation state of oxygen is -2 in all compounds.The value of

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Knowledge Check

  • A_(2)O_(n) is oxidised to AO_(3)^(-) by KMnO_(4) in acidic medium. If 1.34 m mol of A_(2)O_(n) requires 32.2 ml of 0.05M acidified KMnO_(4) solution for complete oxidation, the correct statemet(s) is/are

    A
    `n =2`
    B
    Empirical formula of the oxide is AO
    C
    1 mole of `A_(2)O_(n)` would require 1 mole of `K_(2)Cr_(2)O_(7)` in acidic medium
    D
    A' can be a IIA group metal

  • y' ml of 'y' volume H_(2)O_(2) solution completely reacts with 'x' ml of 'y' M KMnO_(4) solution in acidic medium . The value of x//y is

    A
    0.036
    B
    28
    C
    0.057
    D
    cannot be calculated

  • The volume of 10 volume H_(2)O_(2) solution that decolourises 200 mL of 2 N KMnO_(4) solution in acidic medium is

    A
    112 mL
    B
    336 mL
    C
    220 mL
    D
    224 mL

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