Question

The percentage of Se in peroxidase anhydrous enzyme is 0.5% by weight (atomic mass = 78.4). Then minimum molecular mass of peroxidase anhydrous enzyme is _______.
(A) 1.568 × 10⁴
(B) 1.568 × 10³
(C) 15.68
(D) 3.136 × 10⁴

Answer 1

answer : option (A) 1.568 × 10⁴

explanation : percentage of Se in peroxidase anhydrous enzyme is 0.5% by weight.

it means, 0.5g of Se present in 100g of peroxidase anhydrous enzyme.

mass of Se = 0.5 g

atomic mass of Se = 78.4 g/mol

so, no of moles = 0.5g/78.4g/mol ≈ 0.006377 mol

as no of mole of Se = no of mole of Peroxidase anhydrous enzyme

or, 0.006377 = mass of enzyme/molecular mass of enzyme

or, 0.006377 = 100/molecular mass of enzyme

or, molecular mass of enzyme = 100/0.006377 ≈ 15,681.3549 g/mol ≈ 1.568 × 10⁴ g/mol

Answer 2

option( A) is correct

$1.568 \times 10 {}^{4}$