A vessel contains two non-reactive gases neon (monoatomic) 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 neon = 20.2 u, and molecular mass of oxygen = 32.0 u.

Options

A. `3/2`
B. `2/3`
C. `1/3`
D. `1/2`

Explanation

To solve the problem, we need to find the ratio of the number of molecules and the mass density of neon and oxygen in the vessel. We will follow these steps:

### Step 1: Understand the Given Information
We have two gases: Neon (Ne) and Oxygen (O₂). The ratio of their partial pressures is given as:
\[ P_{Ne} : P_{O_2} = 3 : 2 \]

### Step 2: Express Partial Pressures
Let the total pressure of the system be \( P \). From the ratio of partial pressures, we can express the partial pressures of Neon and Oxygen as:
\[ P_{Ne} = \frac{3}{5}P \quad \text{and} \quad P_{O_2} = \frac{2}{5}P \]

### Step 3: Relate Number of Molecules to Partial Pressure
The number of molecules of a gas can be related to its partial pressure using the ideal gas law. For two gases at the same volume and temperature, the ratio of the number of molecules \( n_1 \) (for Neon) to \( n_2 \) (for Oxygen) is given by:
\[ \frac{n_{Ne}}{n_{O_2}} = \frac{P_{Ne}}{P_{O_2}} \]
Substituting the values we have:
\[ \frac{n_{Ne}}{n_{O_2}} = \frac{3/5 P}{2/5 P} = \frac{3}{2} \]

### Step 4: Calculate the Ratio of Mass Densities
The density \( \rho \) of a gas can be expressed as:
\[ \rho = \frac{m}{V} \]
where \( m \) is the mass and \( V \) is the volume. The mass can also be expressed in terms of the number of moles and molar mass:
\[ m = n \cdot M \]
where \( n \) is the number of moles and \( M \) is the molar mass. Thus, the density can be rewritten as:
\[ \rho = \frac{n \cdot M}{V} \]
For both gases, since the volume is the same, the ratio of densities becomes:
\[ \frac{\rho_{Ne}}{\rho_{O_2}} = \frac{n_{Ne} \cdot M_{Ne}}{n_{O_2} \cdot M_{O_2}} \]
Substituting the values we have:
\[ \frac{\rho_{Ne}}{\rho_{O_2}} = \frac{\left( \frac{3}{2} n_{O_2} \right) \cdot 20.2}{n_{O_2} \cdot 32.0} \]
This simplifies to:
\[ \frac{\rho_{Ne}}{\rho_{O_2}} = \frac{3 \cdot 20.2}{2 \cdot 32.0} = \frac{60.6}{64.0} \approx 0.946875 \]

### Step 5: Final Answers
Thus, the final ratios are:
1. The ratio of the number of molecules of Neon to Oxygen is \( \frac{3}{2} \).
2. The ratio of the mass densities of Neon to Oxygen is approximately \( 0.946875 \).