Question

Total number of atoms present in $34\, g$ of $NH_3$ is

• A. $4\times10^{23}$
• B. $4.8\times10^{21}$
• C. $2\times10^{23}$
• D. $48\times10^{23}$

Answer 1

Answer: (D)

No. of moles of $34 \,g$ of $NH_3 = \frac {34}{17} = 2$
No. of molecules $= 2 \times 6.023 \times 10^{23}$
No. of atoms in one molecule of $NH_3 = 4$
No. of atoms in $2 \times 6.023 \times 10^{23}$ molecules of $NH_{3}$
$ = 4\times 2 \times 6.023 \times 10^{23}$
$ = 48.18 \times 10^{23}$