Question

$1cc{N}_{2}O$ 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 $22400cc$ of $N_{2}O$ contains $=6.02\times 10^{23}$ molecules $\therefore 1cc{N}_{2}O$ 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