Question
What will be the weight of $$CO$$ having the same number of oxygen atoms as present in $$22\,g$$ of $$C{O_2}?$$
A.
28$$\,g$$
B.
22$$\,g$$
C.
44$$\,g$$
D.
72$$\,g$$
Answer :
28$$\,g$$
Solution :
$$\eqalign{
& {\text{No}}{\text{. of}}\,\,O\,\,{\text{atoms}}\,\,{\text{in}}\,\,C{O_2} = 2 \cr
& {\text{Molar mass of}}\,\,C{O_2} = 44\,g \cr
& 44\,\,g \equiv 1\,mol \Rightarrow 22\,g \equiv 0.5\,mol \cr} $$
$$1\,mole\,\,{\text{of}}\,\,C{O_2}\,\,{\text{contains}}$$ $$ = 2 \times 6.023 \times {10^{23}}O\,\,{\text{atoms}}$$
$$0.5\,mole\,\,{\text{of}}\,\,C{O_2}\,\,{\text{contains}}$$ $$= 6.023 \times {10^{23}}O\,\,{\text{atoms}}$$
$$1\,mole\,\,{\text{of}}\,\,CO\,\,{\text{contains}}$$ $$ = 6.023 \times {10^{23}}O\,\,{\text{atoms}}$$
$${\text{Mass of}}\,\,1\,mole\,\,{\text{of}}\,\,CO$$ $$ = 12 + 16 = 28\,g$$