Question

Percentage of Se in peroxide anhydrous enzyme is 0.5% by weight (At. wt. $=78.4)$ then miniumum molecular weight of peroxidase anhydrous enzyme is

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

Answer 1

The correct option is A $1.568\times {10}^4$

Let ${}^{\prime }{M}^{\prime }gmo{l}^{-1}$ be the minimum molecular weight of peroxidase anhydrous enzyme.
Minimum number of atoms of $Se$ present in one molecule of peroxidase anhydrous enzyme is one.
Minimum number of moles of $Se$ present in one mole of peroxidase anhydrous enzyme is one.
Hence, $78.4g$ of $Se$ is present in $Mg$ of peroxidase anhydrous enzyme.
Mass percent of $Se=\frac{78.4}{M}\times 100=\frac{7840}{M}\%$
Given the percentage of $Se$ in peroxidase anhydrous enzyme is $0.5\%$ by weight:
$\therefore \frac{7840}{M}=0.5$
$\Rightarrow M=\frac{7840}{0.5}$
Thus, $M=1.568\times {10}^{4}gmo{l}^{-1}$

Hence the correct answer is option (a).