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Percentage of Se in peroxidase anhydrase...

Percentage of Se in peroxidase anhydrase enzyme is `0.5%` by weight (at. Wt. `=78.4)`, then minimum molecular weight of peroxidase anhydrase enzyme is:

A

`1.568xx10^(4)`

B

`1.568xx10^(3)`

C

`15.68`

D

`3.136xx10^(4)`

Text Solution

Verified by Experts

The correct Answer is:
A
`0.5%` by weight means 0.5 g Se is present in 100 g of peroxidase anhydrous enzyme. As at least one atom of Se must be present in the enzyme and atomic weight of Se = 78.4, therefore, 1 g atom of Se, i.e., 78.4 g will be present in enzyme
`=(100)/(0.5)xx78.4g=1.568xx10^(4)g`
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Knowledge Check

  • The percentage of Se in peroxidase enzyme is 0.5% by weight (atomic weight = 78.4). Then minimum molecular weight of peroxidase anhydrous enzyme is

    A
    `1.568 xx 10^(4)`
    B
    `1.568 xx 10^(3)`
    C
    `15.68`
    D
    `3.136 xx 10^(4)`

  • Percentage of Se (at. mass=78.4) in peroxidase anhydrase enzyme is 0.5% by weight, then minimum molecular mass of peroxidase anhydrase enzyme is

    A
    `1.576 xx 10^4`
    B
    `1.576 xx 10^3`
    C
    `15.76`
    D
    `2.136 xx 10^4`

  • Percentage of Se (at. mass 3 78.4) in peroxidase anhydrase enzyme is 0.5% by weight, then minimum molecular mass of peroxidase anhydrase enzyme is

    A
    `1.568xx10^(4)`
    B
    `1.568xx10^(3)`
    C
    `15.68`
    D
    `2.136 xx 10^(4)`

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