Question

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

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

Answer 1

Answer: (A), (B)

14 g of $N_2$

Calculating the number of molecules,
using the given information in the options:

No. of moles $=\frac{Givenmass}{Molecularmass}$

No. of molecules $=Avogadr{o}^{\prime }sno.\times no.ofmoles$

For option (A)

Molecular mass = 16+12 = 28 g/mol
Number of molecules in 16g of CO $=\left(16/28\right)\times 6.023\times {10}^{23}$
$=0.571\times 6.023\times {10}^{23}$

For option (B)

28 g of $N_2$
Molecular mass = 14+14 = 28 g/mol
Number of molecules in 28g of $N_2=\left(28/28\right)\times 6.023\times {10}^{23}$
$=6.023\times {10}^{23}$


For option (C)

14g of $N_2$

Number of molecules in 14g of $N_2=\left(14/28\right)\times 6.023\times {10}^{23}$
$=0.5\times 6.023\times {10}^{23}$

For option (D)

1.0 g of $H_2$
Molecular mass of $H_2$ = 1+1 = 2 g/mol
Number of molecules in 1.0 g of $H_2=\left(1/2\right)\times 6.023\times {10}^{23}$
$=0.5\times 6.023\times {10}^{23}$

16 g of $O_2$
Molecular mass = 16+16 = 32 g/mol
Number of molecules in 16 g of $O_2=\left(16/32\right)\times 6.023\times {10}^{23}$
$=0.5\times 6.023\times {10}^{23}$

Hence from the above calculation, it is clear that option (C) and (D) have the same number of molecules as 16 g of oxygen has. i.e. $=0.5\times 6.023\times {10}^{23}$