Question

The system that contains the maximum number of atoms is

• A. $4.25g$ of $N{H}_3$
• B. $8g$ of $O_2$
• C. $2g$ of $H_2$
• D. $4g$ of $He$

Answer 1

Answer: (C)

Number of atoms $=\frac{\text{ mass }}{\text{ molar mass }}\times N_A$
$\times$ Number of atoms in $1$ mole
$\therefore$ Number of atoms in $4.25gN{H}_3$
$=\frac{4.25}{17}\times N_A\times 4$
$=N_A$
Number of atoms in $8g$
$O_2=\frac{8}{32}\times N_A\times 2=\frac{N_A}{2}$
Number of atoms in $2g$
$H_2=\frac{2}{2}\times N_A\times 2=2N_A$
Number of atoms in $4g$
$He=\frac{4}{4}\times N_A\times 1=N_A$
Thus, $2g{H}_2$ contains the maximum number of atoms among the given.