Single Choice

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$$\displaystyle 1.568\times 10^{4}$$
Correct Answer
B$$\displaystyle 1.568\times 10^{3}$$
C$$15.68$$
D$$\displaystyle 2.136\times 10^{4}$$

Solution

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 $$\displaystyle \frac{100}{0.5}\times 78.4 = $$Molecular weight

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