Question

The number of oxygen atoms in $4.4g$ of $C{O}_2$ is,

• A. $1.2\times 10^{23}$
• B. $6\times 10^{22}$
• C. $6\times 10^{23}$
• D. $12\times 10^{23}$

Answer 1

Answer: (A)

Molar mass of $C{O}_2=12+32=44$

$\because$ $\frac{W}{M}=\frac{N}{N_A}$ (for $C{O}_2)$

$\frac{4.4}{44}=\frac{N}{6\times 10^{23}}$

$\therefore N=\frac{4.4\times 6\times 10^{23}}{44}$

$=0.1\times 6\times 10^{23}=6\times 10^{22}$

$N=2$ times of $6\times 10^{22}$

$(\because$ Each $C{O}_2$ contains two oxygen atoms)

$\therefore N$ (oxygen) $=2\times 6\times 10^{22}$

$=1.2\times 10^{23}$