Question

percentage of Se in peroxidase anhydrous enzyme is 0.5 % by weight (at wt=78.4) then minimum molecular weight of peroxidase anhydrous enzyme is

Answer 1

let molecular weight of enzyme be x
weight of Se in enzyme = 0.5% of x
= 0.5 x / 100
at. wt. of Se = 78.4 (which is given in the question)
so,
78.4=0.5 x / 100
78.4 * 100 = 0.5 x
7840 = 0.5 x
x = 7840 / 0.5
x = 7840 * 10 / 5

therefore , x = 15680 or 1.5680 * 10 to the power 4

Answer 2

Answer:

i hope this help

the first answer is also same