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

(a) 1.568 × 10⁴

(b) 1.568 × 10³

(c) 15.68

(d) 2.136 × 10⁴

Answer 1

Ans. (a) 1.568 × 10⁴

Sol. Given -

0.5% of Se

that is there is 0.5g of Se in 100g of peroxidase anhydrous enzyme

to find the minimum molecular weight of enzyme we consider the minimum amount of Selenium in the enzyme to be is one atom (its atomic mass)

let the minimum molecular wt of enzyme be x

mass  of Se	mass of enzume
0.5g	100g
78.4g	xg

now, for 0.5g of Se -> 100g of enzyme 

         for 1g of Se -> 100 / 0.5 = 200g of enzyme 

thus for 78.4g of Se -> (200 x 78.4)g = 15680g of enzyme 

MINIMUM MOLECULAR WEIGHT OF PEROXIDASE ANHYDROUS ENZYME = 15680g = 1.568 x 10⁴ g