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In laboratory K(2)Cr(2)O(7) is used main...

In laboratory `K_(2)Cr_(2)O_(7)` is used mainly not `Na_(2)Cr_(2)O_(7).` Why?

Text Solution

Verified by Experts

`Na_(2)Cr_(2)O_(7)` is deliquescent enough and changes its concetration and can not be taken as primary standard solution whereas `K_(2)Cr_(2)O_(7)` has no water of crystallisation and not deliquescent.
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An excess KI solution is mixed in a solution of K_(2)Cr_(2)O_(7) and liberated iodine required 72 mL of 0.05 N Na_(2)S_(2)O_(3) for complete reaction. How many grams of K_(2)Cr_(2)O_(7) were present in the solution of K_(2)Cr_(2)O_(7) ? The reaction occurs as : Cr_(2)O_(7)^(2-)+6I^(-)+14H^(+) to 2Cr^(3+)+3I_(2)+7H_(2)O

K_(2)Cr_(2)O_(7) , on heating gives

Knowledge Check

  • K_(2)Cr_(2)O_(7) is an example of

    A
    hexagonal
    B
    triclinic
    C
    cubic
    D
    Orthorhombic.

  • K_(2)Cr_(2)O_(7) is preferred oxidizing agent than Na_(2)Cr_(2)O_(7) because

    A
    `Na_(2)Cr_(2)O_(7)` is hygroscopic
    B
    `K_(2)Cr_(2)O_(7)` is hygroscopic
    C
    `Na_(2)Cr_(2)O_(7)` is not an oxidizing agent
    D
    `K_(2)Cr_(2)O_(7)` is oxidizing and hygroscopic

  • K_(2)Cr_(2)O_(7) , on heating gives

    A
    `Cr_(2)O_(3)`
    B
    `K_(2)CrO_(4)`
    C
    `O_(2)`
    D
    All of these

    • Similar Questions

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    Statement-1: K_(2)Cr_(2)O_(7)+2NaCl toNa_(2)Cr_(2)O_(7)+2KCl Statement-2: K_(2)Cr_(2)O_(7) is less soluble than Na_(2)Cr_(2)O_(7) .

    Acidic K_(2) Cr_(2)O_(7) oxidises KI to

    Assertion (A): Potassium dichromate is preferred to Na_(2)Cr_(2)O_(7) for use in volumetric analysis (titrations) Reason (R): Na_(2)Cr_(2)O_(7) is hygroscopic whilst the K_(2)Cr_(2)O_(7) is not Which of the following is correct ?

    In K_(2)Cr_(2)O_(7) , every Cr is linked to

    In the standardisation of Na_(2)S_(2)O_(3) using K_(2)Cr_(2)O_(7) by iodometry the equivalent mass of K_(2)Cr_(2)O_(7) is:

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