Question
Total number of atoms present in $$34\,g$$ of $$N{H_3}$$ is
A.
$$4 \times {10^{23}}$$
B.
$$4.8 \times {10^{21}}$$
C.
$$2 \times {10^{23}}$$
D.
$$48 \times {10^{23}}$$
Answer :
$$48 \times {10^{23}}$$
Solution :
No. of moles of $$34\,g$$ of $$N{H_3} = \frac{{34}}{{17}} = 2$$
No. of molecules $$ = 2 \times 6.023 \times {10^{23}}$$
No. of atoms in one molecule of $$N{H_3} = 4$$
No. of atoms in $$2 \times 6.023 \times {10^{23}}$$ molecules of $$N{H_3}$$
$$\eqalign{
& = 4 \times 2 \times 6.023 \times {10^{23}} \cr
& = 48.18 \times {10^{23}} \cr} $$