Question

Enzyme contains 5.6 % iron. The minimum molecular mass of the enzyme is

Answer 1

Answer:

Answer : The number of Fe atoms present in 1 g of enzyme is,

Solution : Given,

An enzyme contains 5.6 % Fe that means, in 100 g of enzyme 5.6 g of Fe is present.

As, 100 g of enzyme contains 5.6 g of Fe

So, 1 g of enzyme contains g of Fe

Now we have to calculate the number of Fe atoms.

1 mole weighs 56 grams as contains number of atoms

As, 56 grams contains number of Fe atoms

So, 0.056 grams contains number of Fe atoms

Answer 2

Answer:

The minimum molecular mass of the enzyme is$6.022 \times 10^{20}$

Explanation:

Mass of $\mathrm{Fe}$is given so will take mass percentage:

Given weight of enzyme $=1 \mathrm{~g}$

$\therefore \quad Weight of \mathrm{Fe} atoms =5.6 \% of 1 \mathrm{~g}$

$\begin{aligned}&=\frac{56}{100} \times 1 \\&=0.056 \mathrm{~g} \text { of } \mathrm{Fe} \text { atoms }\end{aligned}$

1 mole of Fe atoms weighs $56 \mathrm{~g}$

Number of Fe atoms present in $1 \mathrm{~g}$ of

$\begin{aligned}\text { Enzyme } &=\frac{\text { mass }}{\text { molecular mass }} \times N_{A} \\&=\frac{0.056}{56} \times 6.022 \times 10^{23} \\&=6.022 \times 10^{20}\end{aligned}$