
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
581.1k+ views
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.
Recently Updated Pages
Why are manures considered better than fertilizers class 11 biology CBSE

Find the coordinates of the midpoint of the line segment class 11 maths CBSE

Distinguish between static friction limiting friction class 11 physics CBSE

The Chairman of the constituent Assembly was A Jawaharlal class 11 social science CBSE

The first National Commission on Labour NCL submitted class 11 social science CBSE

Number of all subshell of n + l 7 is A 4 B 5 C 6 D class 11 chemistry CBSE

Trending doubts
Differentiate between an exothermic and an endothermic class 11 chemistry CBSE

10 examples of friction in our daily life

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

State the laws of reflection of light

