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)

All of the above

Answer 2

At NTP 22400 cc of $N_2O$ contains = $6.02 \times 10^{23} molecules$
$\therefore 1 cc \, N_2O \, will \, contain \, = \frac {6.02 \times 10^{23}}{22400}molecules$
In $N_2O$ 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 $N_2O$ molecule, number of electrons
$\, \, \, \, \, \, \, \, \, \, \, \, \, = 7 + 7 + 8 = 22$
Hence, number of electrons $\frac {6.02 \times 10^{23}}{22400} \times 22$ electrons
$= \frac {1.32 \times 10^{23}}{224} electrons$