Question

The ratio of number of oxygen atoms (O) in 16.0 g ozone $(O_3), \,28.0\, g$ carbon monoxide $(CO)$ and $16.0$ oxygen $(O_2)$ is (Atomic mass: $C = 12,0 = 16$ and Avogadro’s constant $N_A = 6.0 x 10^{23}\, mol^{-1}$)

• A. $3 : 1 : 2$
• B. $1 : 1 : 2$
• C. $3 : 1 : 1$
• D. $1 : 1 : 1$

Answer 1

Answer: (D)

$O _{3}$ molecular weight $=16+16+16=48 g / mol$
It means weight of 1 mol of $O _{3}$ is $48 g$ and in $1 mol$ of $O _{3}$ we have 3 atoms of Oxygen
In $48 g$ of $O _{3}$, number of atoms of oxygen $=3$
so in $16 g$ of $O _{3}$, number of atoms of oxygen $=(3 / 48) \times 16=1$
CO molecular weight $=12+16=28 g / mol$
It means weight of $1 mol$ of $CO$ is $28 g$ and in $1 mol$ of $CO$ we have 1 atom of Oxygen
so in $28 g$ of $CO$, number of atoms of oxygen $=1$
O2 molecular weight $=12+16=32 g / mol$
It means weight of $1 mol$ of $O _{2}$ is $32 g$ and in $1 mol$ of $O _{2}$ we have 2 atoms of Oxygen
In $32 g$ of $O _{2}$, number of atoms of oxygen $=2$
so in $16 g$ of $O _{2}$, number of atoms of oxygen $=(2 / 32) \times 16=1$
So answer is $1: 1: 1$