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Chemistry
1 cc N2O at NTP contains
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Q. $1 \,cc\, N_2O$ at NTP contains
AIPMT
AIPMT 1988
Some Basic Concepts of Chemistry
A
$\frac {1.8}{224} \times 10^{22}$ atoms
9%
B
$\frac {6.02}{22400} \times 10^{23} $ molecules
26%
C
$\frac {1.32}{224} \times 10^{23}$ electrons
11%
D
All of the above
54%
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