Question # Percentage of Se in peroxidase enzyme is ( 0.5 % ) by weight ( (mathrm{Se}=78.4) ). Then the minimum molecular weight of peroxidase enzyme is

# Percentage of Se in peroxidase enzyme is ( 0.5 % ) by weight ( (mathrm{Se}=78.4) ). Then the minimum molecular weight of peroxidase enzyme is

(1) ( 3.13 times 10^{4} )

( begin{array}{ll}text { (2) } & 15.68end{array} )

(3) ( 1.568 times 10^{3} )

( begin{array}{ll}text { (4) } 1.568 times 10^{4} & text { 4 }end{array} )

Solution

Minimum molecular weight of peroxidase anhydrous enzyme is ( 15,792 mathrm{g} / mathrm{mol} ).

Explanation:

Let the molar mass of the peroxidase anhydrous be M

One molecule of peroxidase anhydrous contain 1 atom of Se

( 0.5 %=frac{78.96 g / m o l}{M} times 100 )

( M=frac{78.96 mathrm{g} / mathrm{mol}}{0.5} times 100=15,792 mathrm{g} / mathrm{mol} )

Minimum molecular weight of peroxidase anhydrous enzyme is ( 15,792 mathrm{g} / mathrm{mol} )