Question

Haemoglobin contains $0.34\%$ of iron by mass. The number of Fe atoms in $3.3$ g of haemoglobin is(Given: Atomic mass of Fe is $56$ u, $N_A$ = $6.022 × 1023$ $mol^{–1}$)

• A. $1.21 × 10^5$
• B. $12.0 × 10^{16}$
• C. $1.21 × 10^{20}$
• D. $3.4 × 10^{22}$

Answer 1

Answer: (C)

$1.21 × 10^{20}$

Answer 2

$100$ g of haemoglobin contains $0.34$ g of iron (As per the question)

$3.3$ g of haemoglobin contains

$\frac{0.34}{100}×3.3$ g of iron moles of $Fe$

=$\frac{0.34×3.3}{100×56} = \frac{N}{N_A}$

$N =\frac{ 0.34 \times 3.3 \times 6.022 \times 10^{23}}{100 \times 56}$

= $1.21 \times 10^{20}$