Question

$1 \,cc\, N_2O$ at NTP contains

• A. $\frac {1.8}{224} \times 10^{22}$ atoms
• B. $\frac {6.02}{22400} \times 10^{23} $ molecules
• C. $\frac {1.32}{224} \times 10^{23}$ electrons
• D. All of the above

Answer 1

Answer: (D)

At NTP $22400\, cc$ of $N _{2} O$ contains $=6.02 \times 10^{23}$ molecules $\therefore \,1 \,cc \,N _{2} O$ will contain $=\frac{6.02 \times 10^{23}}{22400}$ molecules
In $N _{2} O$ molecule, number of atoms $=2+1=3$
Thus, number of atoms $=\frac{3 \times 6.02 \times 10^{23}}{22400}$ atoms
$=\frac{1.8 \times 10^{22}}{224}$ atoms
In $N _{2} O$ molecule, number of electrons
$=7+7+8=22$
Hence, number of electrons $=\frac{6.02 \times 10^{23}}{22400} \times 22$ electrons
$=\frac{1.32 \times 10^{23}}{224}$ electrons