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Question: The percentage of \[Se\] in peroxidase anhydrous enzyme is \(0.5\% \) by weight \((At.\;wt. = 78.4)\...

The percentage of SeSe in peroxidase anhydrous enzyme is 0.5%0.5\% by weight (At.  wt.=78.4)(At.\;wt. = 78.4) then, what would be the minimum molecular weight of peroxidase anhydrous enzyme?
A) 1.568×1041.568 \times {10^4}
B) 1.568×1031.568 \times {10^3}
C) 15.6815.68
D) 2.136×1042.136 \times {10^4}

Explanation

Solution

As we know that mass percentage is basically the ratio where given mass of the substance is present in 100g100g of total mass of the compound and we also know that the moles is the ratio of the given mass of the substance or solute to the molecular mass of the solvent.

Formula used: Mass  %=given  mass  of  atommolecular  mass  of  compound×100Mass\;\% = \dfrac{{given\;mass\;of\;atom}}{{molecular\;mass\;of\;compound}} \times 100

Complete answer:
We know that mass percent of an element is the percentage or composition of a component in a compound where the percent of the total mass of the compound is due to that component only.
In the question, we are given that Selenium is present in a peroxidase anhydrous enzyme as 0.5g0.5g in 100g100g of enzyme and the atomic mass of selenium is 78.4g78.4g.

So let us consider the molecular mass of the given peroxidase anhydrous enzyme be x  gmol1x\;gmo{l^{ - 1}}.
We know that one mole of peroxidase anhydrous enzyme will contain a minimum of one mole of selenium or we can say that one molecule of peroxidase anhydrous enzyme will contain a minimum of one atom of selenium.

So, we can say that 78.4g78.4g of selenium will be present in x  gmol1x\;gmo{l^{ - 1}}of peroxidase anhydrous enzymes. So the mass percentage can be given as:
Mass  %=given  mass  of  atommolecular  mass  of  compound×100\Rightarrow Mass\;\% = \dfrac{{given\;mass\;of\;atom}}{{molecular\;mass\;of\;compound}} \times 100
Mass  %  of  Se=78.4x×100\Rightarrow Mass\;\% \;of\;Se = \dfrac{{78.4}}{x} \times 100
Mass  %  of  Se=7840x\Rightarrow Mass\;\% \;of\;Se = \dfrac{{7840}}{x}
Now we can calculate the percentage of selenium which is present as 0.5%0.5\% by weight in peroxidase anhydrous enzymes. We will get:
0.5=7840x\Rightarrow 0.5 = \dfrac{{7840}}{x}
x=78400.5=1.568×104  gmol1\Rightarrow x = \dfrac{{7840}}{{0.5}} = 1.568 \times {10^4}\;gmo{l^{ - 1}}

Therefore from the above solution we can say that the correct answer is (A).

Note: Remember that percentage composition of the compound is the relative mass of the each of the constituent element in 100100 parts of it or mass percent of an element is the ratio of mass of that element in one mole of the compound to the molar mass of the compound multiplied by 100100. We can also check the purity of a given sample by analysing the percentage composition.