Question:
The number of atoms in $8 \mathrm{~g}$ of sodium is $\mathrm{x} \times 10^{23}$.
The value of $x$ is .(Nearest integer)
[Given : $\mathrm{N}_{\mathrm{A}}=6.02 \times 10^{23} \mathrm{~mol}^{-1}$
Atomic mass of $\mathrm{Na}=23.0 \mathrm{u}]$
Solution:
No. of atoms $=\frac{8}{23} \times 6.02 \times 10^{23}=2.09 \times 10^{23}$
$\simeq 2 \times 10^{23}$
$=x \times 10^{23}$
$x=2$
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.