Question

The number of oxygen atoms in $4.4\, g$ of $CO_2$ 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 $CO _{2}=12+32=44$

$\because$ $\frac{W}{M} =\frac{N}{N_{A}}$ (for $CO _{2})$

$ \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 $CO _{2}$ contains two oxygen atoms)

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

$=1.2 \times 10^{23}$