1.

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

Answer»

`1.568 xx 10^(4)`
`1.568 xx 10^(3)`
`15.68`
`3.136 xx 10^(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:
`=100/0.5 xx 78.4= 1.568 xx 10^(4)` g in enzyme.


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