Question

An enzyme contains 5.6% fe calculate number of fe atoms present in 1 g of enzyme

Answer 1

The mass of Fe present= 5.6% of 1g = 5.6 ×10^(-2) g

Answer 2

Answer : The number of Fe atoms present in 1 g of enzyme is, $6.022\times 10^{20}$

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 $\frac{5.6}{100}=0.056$ g of Fe

Now we have to calculate the number of Fe atoms.

1 mole weighs 56 grams as contains $6.022\times 10^{23}$ number of atoms

As, 56 grams contains $6.022\times 10^{23}$ number of Fe atoms

So, 0.056 grams contains $\frac{0.056}{56}\times (6.022\times 10^{23})=6.022\times 10^{20}$ number of Fe atoms

Therefore, the number of Fe atoms present in 1 g of enzyme is, $6.022\times 10^{20}$