Question

The percentage of $Se$ in peroxidase anhydrous enzyme is $0.5$% by weight (atomic weight $=78.4$). Then minimum molecular weight of peroxidase anhydrous enzyme is:

• A. $1.568\times {10}^4$
• B. $1.568\times {10}^3$
• C. $15.68$
• D. $3.136\times {10}^4$

Answer 1

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

Let $M$ g/mol 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, 78.4 g of $Se$ are present in M g of peroxidase anhydrous enzyme.

Mass percent of $Se$ $=\frac{78.4g}{Mg}\times 100=\frac{7840}{M}$

The percentage of $Se$ in peroxidase anhydrous enzyme is $0.5$% by weight:

$\frac{7840}{M}=0.5$

$M=\frac{7840}{0.5}$

$M=1.568\times {10}^4$ g/mol