Q. The number of atoms present in a 0.635 g of Cu piece will be
A) $6.023 \times 10^{- 23}$
B) $6.023 \times 10^{23}$
C) $6.023 \times 10^{22}$
D) $6.023 \times 10^{21}$
Explanation
(d)
$\because$ 63.8 g of Cu has atoms = 6.023 $\times 10^{23}$
$\therefore 1g\$of Cu has = $\frac{6.023 \times 10^{23}}{63.5g}$
$\therefore 0.635\ g$ of Cu has $= \frac{6.023 \times 10^{23}}{63.5} \times 0.635$
$= 6.023 \times 10^{21}$ atoms
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.