$A$ vessel contains $1.6 \ g$ of dioxygen at $STP$ ($273.15 \ K$,$1 \ atm$ pressure). The gas is now transferred to another vessel at constant temperature,where pressure becomes half of the original pressure. Calculate
$(A)$ volume of the new vessel.
$(B)$ number of molecules of dioxygen.

(N/A) Step $1$: Calculate the number of moles of $O_2$.
Given mass of $O_2 = 1.6 \ g$.
Molar mass of $O_2 = 32 \ g/mol$.
Moles $(n)$ = $\frac{1.6 \ g}{32 \ g/mol} = 0.05 \ mol$.
Step $2$: Calculate the initial volume $(V_1)$ at $STP$.
At $STP$,$1 \ mol$ of gas occupies $22.7 \ L$ (using $1 \ bar$ pressure) or $22.4 \ L$ (using $1 \ atm$ pressure).
Using $P_1 = 1 \ atm$,$V_1 = n \times 22.4 \ L/mol = 0.05 \ mol \times 22.4 \ L/mol = 1.12 \ L$.
Step $3$: Calculate the volume of the new vessel $(V_2)$.
Since temperature is constant,we use Boyle's Law: $P_1 V_1 = P_2 V_2$.
Given $P_2 = \frac{1}{2} P_1$,so $P_1 V_1 = (\frac{1}{2} P_1) V_2$.
$V_2 = 2 V_1 = 2 \times 1.12 \ L = 2.24 \ L$.
Step $4$: Calculate the number of molecules.
Number of molecules = $n \times N_A$.
$= 0.05 \ mol \times 6.022 \times 10^{23} \ molecules/mol = 3.011 \times 10^{22} \ molecules$.