Question
The largest number of molecules is in
A.
$$36\,g$$ of water
B.
$$28\,g$$ of carbon monoxide
C.
$$46\,g$$ of ethyl alcohol
D.
$$54\,g$$ of nitrogen pentoxide
Answer :
$$36\,g$$ of water
Solution :
$$\eqalign{
& {\text{(A) }}18\,g\,{\text{of}}\,{H_2}O = 6.02 \times {10^{23}}{\text{molecules}}\,{\text{of}}\,{H_2}O \cr
& \therefore 36\,g\,{\text{of }}{H_2}O = 2 \times 6.02 \times {10^{23}}{\text{molecules}}\,{\text{of}}\,{H_2}O \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{ = }}12.04 \times {10^{23}}{\text{molecules}}\,{\text{of}}\,{H_2}O \cr
& {\text{(B)}}\,\,{\text{28}}\,{\text{g}}\,{\text{of}}\,CO = 6.02 \times {10^{23}}{\text{molecules}}\,{\text{of}}\,CO \cr
& {\text{(C)}}\,\,46\,g{\text{ of}}\,{C_2}{H_5}OH = 6.02 \times {10^{23}}{\text{molecules}}\,{\text{of}}\,{C_2}{H_5}OH \cr
& ({\text{D}})\,\,108\,\,g\,\,{\text{of}}\,{N_2}{O_5} = 6.02 \times {10^{23}}{\text{molecules}}\,{\text{of}}\,{N_2}{O_5} \cr
& \therefore 54\,g\,\,{\text{of}}\,{N_2}{O_5} = \frac{1}{2} \times 6.02 \times {10^{23}}{\text{molecules}}\,{\text{of}}\,{N_2}{O_5} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 3.01 \times {10^{23}}{\text{molecules}}\,{\text{of}}\,{N_2}{O_5} \cr
& \therefore 36\,g\,{\text{of}}\,{\text{water}}\,{\text{has}}\,{\text{highest}}\,{\text{number}}\,{\text{of}}\,{\text{molecules}}{\text{.}} \cr} $$