Question
How many oxygen atoms will be present in $$88\,g$$ of $$C{O_2}?$$
A.
$$24.09 \times {10^{23}}$$
B.
$$6.023 \times {10^{23}}$$
C.
$$44 \times {10^{23}}$$
D.
$$22 \times {10^{24}}$$
Answer :
$$24.09 \times {10^{23}}$$
Solution :
$$\eqalign{
& 1\,mole\,\,{\text{of}}\,\,C{O_2} = 44\,g \cr
& 88\,g\,\,{\text{of}}\,\,C{O_2} = 2\,moles \cr
& {\text{No}}{\text{. of oxygen atoms in}}\,\,2\,moles \cr
& = 2 \times 2 \times {N_A} \cr
& = 2 \times 2 \times 6.023 \times {10^{23}} \cr
& = 24.09 \times {10^{23}} \cr} $$