1.

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»

: 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 gatomic mass of Se = 78.4 g/molso, no of moles = 0.5g/78.4g/MOL ≈ 0.006377 mol as no of mole of Se = no of mole of Peroxidase anhydrous enzymeor, 0.006377 = mass of enzyme/molecular mass of enzymeor, 0.006377 = 100/molecular mass of enzymeor, molecular mass of enzyme = 100/0.006377 ≈ 15,681.3549 g/mol ≈ 1.568 × 10⁴ g/mol



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