Question

Percentage of Se in peroxidase anhydrous enzyme is $0.5$% by weight (At. wt. = $78.4$). What is the minimum molecular weight of peroxidase anhydrous enzyme?
A. $1.568 \times \,{10^4}$
B. $1.568 \times \,{10^3}$
C. $15.68$
D. $2.136 \times \,{10^4}$

Answer 1

Hint:We will assume the total amount of enzyme as$100$. The amount of Se in the assumed amount is given. By comparing the stoichiometry we can calculate the amount of peroxidase anhydrous enzyme in which $78.4$gram Se is present.

Complete step-by-step solution
The mass equal to the one-twelve of the mass of carbon-$12$ is known as atomic mass. The unit of atomic mass is a.m.u. Molecular weight is the mass of a molecule which is calculated by adding the mass of each atom of the molecule.

It is given that of $0.5$% by weight Se is present in peroxidase anhydrous enzymes. It means if the mass of peroxidase anhydrous enzyme is $100$gram than out of$100$gram, $0.5$gram is Se. The atomic weight of Se is$78.4$gram.

So, we can calculate the amount of peroxidase anhydrous enzyme in which$78.4$gram Se will be present as follows:

$0.5$gram of Se =$100$gram peroxidase anhydrous enzyme
$78.4$gram Se =$1.568 \times \,{10^4}$gram
So, $78.4$gram Se will be present in $1.568 \times \,{10^4}$gram of peroxidase anhydrous enzyme.
So, the minimum molecular weight of peroxidase anhydrous enzyme is $1.568 \times \,{10^4}$.

Therefore, option (A) $1.568 \times \,{10^4}$ is correct.

Note:Stoichiometry measurements are used to determine the amount of reactant or product from the given amounts. Stoichiometry measurements give quantitative relations among the amounts of various species of a reaction. X by weight means if the weight of one total substance is hundred then the amount of one component is x. when the by weight relation is given in percentage we assume the total amount as a hundred percent.