1.

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

Answer»

`1.568xx10^(4)`
`1.568xx10^(3)`
`15.68`
`3.136xx10^(4)`

SOLUTION :`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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