SOLUTION : Calculation of number of moles
Volume of the diatomic gas at S.T.P.
$P_{1}=1 \cdot 0 atm \quad P_{2}=2 \cdot 0 atm \quad T_{1}=273 K \quad T_{2}=273 K V_{1}=? V_{2}=350 mL$
$\because \quad P_{1} V_{1}=P_{2} V_{2} \quad$ (Temperature is constant)
$\therefore \quad V_{1}=\frac{P_{2} V_{2}}{P_{1}}=\frac{2 \times 350}{1 \cdot 0} mL =700 mL$ at $S .$. . P.
$\because \quad 22400 mL$ of a gas at S.T.P. $=1.0$ mole
$\therefore 700 mL \quad. \quad. \quad. \frac{700}{22400}$ mole $=\frac{1}{32}$ mole .
$\therefore \frac{1}{32} \text { mole }=\frac{6 \cdot 02 \times 10^{23}}{32} \text { molecules }$
and this number of molecules weight $1.0 g$. Again, the gas is diatomic, i.e., contains two atoms per molecule. Hence no. of atoms
$=\frac{6.02 \times 10^{23}}{32} \times 2 \text { atoms }$
Now, $\frac{6.02 \times 10^{23}}{16}$ atoms weigh $1.0 g$
$\therefore 1 \cdot 0$ atom will weight $\frac{1 \cdot 0 \times 16}{6 \cdot 02 \times 10^{23}} g$ $= 2 \cdot 65 \times 10^{-23} g$
$\therefore$ No. of molecules $=$ no. of mole $\times 6.02 \times 10^{23}$
