Q. 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) $31.80 \times 10^{28}$
Explanation
(c)
Molecular mass of $Na_{2}CO_{3}$
$= 2 \times 23 + 12 + 3 \times 16 = 106$
$\therefore$ 106 g $Na_{2}CO_{3}$ contains
$= 3 \times 6.023 \times 10^{23}$ oxygen atoms
$\therefore$ 10.6 g of $Na_{2}CO_{3}$ will contain
$= \frac{3 \times 6.023 \times 10^{23}}{106} \times 10.6$
$= 18.069 \times 10^{22}$
$= 1.806 \times 10^{23}$ oxygen atoms
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.