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

AIPMTAIPMT 1988Some Basic Concepts of Chemistry

Solution:

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