Question

Percentage of Se (at. wt. = 78.4) in peroxidase anhydrase enzyme is 0.5% by weight, then minimum molecular weight of peroxidase anhydrase enzyme is

(a) 1·568 x 10³

(b) 1·568 x 10⁴

(c) 15.68

(d) 3·136 x 10⁴

Answer 1

Correct option is (b) 1·568 x 10⁴

0.5% by weight means 0.5 g of Se is present in 100 g of enzyme. Now at least one atom of Se (at wt. 78.4) should be present in the enzyme, in other words 1 g atom of selenium (78.4 g) should be present in the enzyme. Hence

0.5 g 100 of Se = 100 g enzyme

78.4 of of Se = \(\frac{100}{0.5}\) × 78.4 = 1.568 x 10⁴ g