Question

Which of the following at NTP does 1 c.c. $N_{2}O$ contain?

• 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 the above

Answer 1

Answer: (D)

ATNTP 22400 cc of $N_{2}O=6.02\times 10^{23}$ molecules
$\therefore 1cc{N}_{2}O=\frac{6.02\times 10^{23}}{22400}molecules$
$=\frac{3\times 6.02\times 10^{23}}{22400}atoms$
$=\frac{1.8}{224}\times 10^{22}atoms$
No of electrons in a molecule of $N_{2}O$
$=7+7+8=22$
Hence no. of electrons
$=\frac{6.02\times 10^{23}}{22400}\times 22electrons=\frac{1.32\times 10^{23}}{224}$