Question

$16g$ of oxygen has same number of molecules as in

• A. $16g$ of $CO$
• B. $28g$ of $N_2$
• C. $14g$ of $N_2$
• D. $1.0g$ of $H_2$

Answer 1

Answer: (C), (D)

$14g$ of $N_2$
$1.0g$ of $H_2$

Number of molecules $N=n\times N_A$, where, $n$ = number of moles, and $N_A$ is Avogadro's number

Number of moles $n=\frac{mass}{givenmass}=\frac{m}{M}$

So, we can write $N=\frac{m}{M}\times N_A$

For 16 g of oxygen, mumber of molecules $N=\frac{16}{32}\times N_A=\frac{N_A}{2}$

(a) For 16 g of $CO$, mumber of molecules $N=\frac{16}{28}\times N_A=\frac{N_A}{1.75}$

(b) For 28 g of $N_2$, mumber of molecules $N=\frac{28}{28}\times N_A=N_A$

(c) For 14 g of $N_2$, mumber of molecules $N=\frac{14}{28}\times N_A=\frac{N_A}{2}$

(d) For 1 g of $H_2$, mumber of molecules $N=\frac{1}{2}\times N_A=\frac{N_A}{2}$

So, options C and D have same number of molecules as in 16 g oxygen.