Question
The number of atoms in $$4.25\,g$$ of $$N{H_3}$$ is approximately
A.
$$4 \times {10^{23}}$$
B.
$$2 \times {10^{23}}$$
C.
$$1 \times {10^{23}}$$
D.
$$6 \times {10^{23}}$$
Answer :
$$6 \times {10^{23}}$$
Solution :
$$\eqalign{
& {\text{Weight of}}\,N{H_3} = 4.25\,g \cr
& {\text{Number of moles of}} \cr
& N{H_3} = \frac{{{\text{weight}}}}{{{\text{Molecular weight}}}} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{4.25}}{{17}} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\, = 0.25\,mol \cr} $$
Number of molecules in $$0.25\,mole$$ of $$N{H_3}$$
$$ = 0.25 \times 6.023 \times {10^{23}}$$
So, number of atoms
$$\eqalign{
& = 4 \times 0.25 \times 6.023 \times {10^{23}} \cr
& = 6.0 \times {10^{23}} \cr} $$