
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
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Hint: In peroxidase anhydrous enzyme, 0.5% Se is present by weight, it means that 0.5 g Se is present in 100 g of enzyme. Minimum molecular formula is defined as the molecular weight divided by the number of atoms of an element present in a molecule.
Complete answer:
As we know, the minimum molecular formula is the molecular weight which is divided by the number of atoms of an element present in a molecule.
$ Minimum\quad molecular\quad weight\quad =\quad \cfrac { Molecular\quad weight }{ No.\quad of\quad atoms\quad of\quad an\quad element }$
In this question, we will be able to find the minimum molecular weight if we consider that in 1 molecule of the peroxidase anhydrous enzyme only 1 Se atom is present. Therefore, the percentage of Se in peroxidase anhydrous enzyme can be found as:
$Percentage \quad of\quad Se\quad =\quad \cfrac { Atomic\quad weight\quad of\quad Se }{ Min.\quad molecular\quad weight\quad (m) } \quad \times \quad 100$
$\implies m\quad =\quad \cfrac { Atomic\quad weight }{ Percentage \quad of\quad Se } \quad \times \quad 100$
Now, given that Atomic weight of Se = 78.4 and 5 of Se = 0.5. Substituting these values in the above equation, we get,
$m\quad =\quad \cfrac { 78.4 }{ 0.5 } \quad \times \quad 100$
$\implies m\quad =\quad 156.8\quad \times \quad 100\quad =\quad 1.568\quad \times \quad { 10 }^{ 4 }\quad g/mol$
Therefore, the minimum molecular weight of Se in peroxidase anhydrous enzymes is $1.586 \times {10}^{4}$.
So, the correct answer is “Option A”.
Note: Don't confuse between the molecular weight and the minimum molecular weight. Molecular weight is the sum of the atomic weights of individual atoms in a molecule. Whereas, minimum molecular weight is the molecular weight divided by the number of atoms of an element present in a molecule.
Complete answer:
As we know, the minimum molecular formula is the molecular weight which is divided by the number of atoms of an element present in a molecule.
$ Minimum\quad molecular\quad weight\quad =\quad \cfrac { Molecular\quad weight }{ No.\quad of\quad atoms\quad of\quad an\quad element }$
In this question, we will be able to find the minimum molecular weight if we consider that in 1 molecule of the peroxidase anhydrous enzyme only 1 Se atom is present. Therefore, the percentage of Se in peroxidase anhydrous enzyme can be found as:
$Percentage \quad of\quad Se\quad =\quad \cfrac { Atomic\quad weight\quad of\quad Se }{ Min.\quad molecular\quad weight\quad (m) } \quad \times \quad 100$
$\implies m\quad =\quad \cfrac { Atomic\quad weight }{ Percentage \quad of\quad Se } \quad \times \quad 100$
Now, given that Atomic weight of Se = 78.4 and 5 of Se = 0.5. Substituting these values in the above equation, we get,
$m\quad =\quad \cfrac { 78.4 }{ 0.5 } \quad \times \quad 100$
$\implies m\quad =\quad 156.8\quad \times \quad 100\quad =\quad 1.568\quad \times \quad { 10 }^{ 4 }\quad g/mol$
Therefore, the minimum molecular weight of Se in peroxidase anhydrous enzymes is $1.586 \times {10}^{4}$.
So, the correct answer is “Option A”.
Note: Don't confuse between the molecular weight and the minimum molecular weight. Molecular weight is the sum of the atomic weights of individual atoms in a molecule. Whereas, minimum molecular weight is the molecular weight divided by the number of atoms of an element present in a molecule.
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