Question

The ratio of number of oxygen atoms (O) in 16.0 g ozone $(O_3),28.0g$ carbon monoxide $(CO)$ and $16.0$ oxygen $(O_2)$ is (Atomic mass: $C=12,0=16$ and Avogadro’s constant $N_A=6.0x10^{23}mo{l}^{-1}$)

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

Answer 1

Answer: (D)

$O_3$ molecular weight $=16+16+16=48g/mol$
It means weight of 1 mol of $O_3$ is $48g$ and in $1mol$ of $O_3$ we have 3 atoms of Oxygen
In $48g$ of $O_3$, number of atoms of oxygen $=3$
so in $16g$ of $O_3$, number of atoms of oxygen $=(3/48)\times 16=1$
CO molecular weight $=12+16=28g/mol$
It means weight of $1mol$ of $CO$ is $28g$ and in $1mol$ of $CO$ we have 1 atom of Oxygen
so in $28g$ of $CO$, number of atoms of oxygen $=1$
O2 molecular weight $=12+16=32g/mol$
It means weight of $1mol$ of $O_2$ is $32g$ and in $1mol$ of $O_2$ we have 2 atoms of Oxygen
In $32g$ of $O_2$, number of atoms of oxygen $=2$
so in $16g$ of $O_2$, number of atoms of oxygen $=(2/32)\times 16=1$
So answer is $1:1:1$