$A$ vessel contains two non-reactive gases: neon (monatomic) and oxygen (diatomic). The ratio of their partial pressures is $3:2$. Estimate the ratio of
$(i)$ number of molecules and
$(ii)$ mass density of neon and oxygen in the vessel.
Atomic mass of $Ne = 20.2 \; u$,molecular mass of $O_2 = 32.0 \; u$.

(N/A) The partial pressure of a gas in a mixture is the pressure it would exert if it alone occupied the vessel at the same volume and temperature. Since both gases are assumed to be ideal and share the same volume $V$ and temperature $T$,we use the ideal gas law $PV = \mu RT$,where $\mu$ is the number of moles.
For neon $(1)$ and oxygen $(2)$:
$P_1 V = \mu_1 RT$ and $P_2 V = \mu_2 RT$
Thus,$\frac{P_1}{P_2} = \frac{\mu_1}{\mu_2}$. Given $\frac{P_1}{P_2} = \frac{3}{2}$,we have $\frac{\mu_1}{\mu_2} = \frac{3}{2}$.
$(i)$ Since $\mu = \frac{N}{N_A}$,where $N$ is the number of molecules and $N_A$ is Avogadro's number:
$\frac{N_1}{N_2} = \frac{\mu_1}{\mu_2} = \frac{3}{2} = 1.5$.
$(ii)$ Mass density $\rho = \frac{m}{V}$. Since $\mu = \frac{m}{M}$ (where $M$ is molar mass),we have $m = \mu M$.
$\frac{\rho_1}{\rho_2} = \frac{m_1/V}{m_2/V} = \frac{m_1}{m_2} = \frac{\mu_1 M_1}{\mu_2 M_2} = \left(\frac{\mu_1}{\mu_2}\right) \times \left(\frac{M_1}{M_2}\right)$.
Substituting the values: $\frac{\rho_1}{\rho_2} = \frac{3}{2} \times \frac{20.2}{32.0} = 1.5 \times 0.63125 = 0.947$.