Question

Number of molecules present in $4.2$ g $N_2$ is $X\times {10}^{22}$. What is the value of $X$(round off to 1 digit)?

Answer 1

$28g$ of $N_2=6.02\times {10}^{23}$ molecules

$4.2g$ of $N_2=\frac{6.02\times {10}^{23}}{28}\times 4.2$ molecules

$\Rightarrow 9.03\times {10}^{22}$ molecules

Hence, the value of $X=9$.