Question

Number of atoms of oxygen present in $10.6\, g$ of $Na _{2} CO _{3}$ will be

• A. $ 6.02\times 10^{23}$
• B. $ 12.04\times 10^{22}$
• C. $ 1.806\times 10^{23}$
• D. $ 31.80\times 10^{2}$

Answer 1

Answer: (C)

Molecular mass of
$Na_2CO_3 = 2 \times 23 + 12 + 3 \times 16= 106$
$\because 106\, g\,Na_2CO_3$ contains
$= 3 \times 6.023 \times 10^{23}$ oxygen atoms
$\therefore 10.6\, g$ of $Na_2CO_3 $ will contain
$= \frac{3 \times 6.023 \times 10^{23}}{106} \times 10.6$
$= 18.069 \times 10^{24} $
$= 1.806 \times 10^{23}$ oxygen atoms