Question

Percentage of Se in peroxidase anhydrous enzyme is $0.5$% by weight (At. wt. = $78.4$) then, minimum molecular weight of peroxidase anhydrous enzymes is :

• A. $1.568\times {10}^4$
• B. $1.568\times {10}^3$
• C. $15.68$
• D. $2.136\times {10}^4$

Answer 1

Answer: (A)

$1.568\times {10}^4$

In peroxidase anhydrous enzyme 0.5% Se is present means, 0.5 g Se is present in 100 g of enzyme. In a molecule of enzyme one Se atom must be present. Hence 78.4 g Se will be present :

$0.5$ g Se present in $100$ g enzyme.

$\therefore 78.4$ g present in $\frac{100}{0.5}\times 78.4=$Molecular weight