At 273 K, the density of a certain gaseous oxide at 2 atmosphere is same as that of dioxygen at 5 atmosphere. The molecular mass of the oxide `("in g mol"^(-1))` is

Options

A. 80
B. 64
C. 70
D. 160

Explanation

To find the molecular mass of the gaseous oxide, we can start by using the relationship between density, pressure, and molecular mass. The formula for density (d) is given by:

\[ d = \frac{PM}{RT} \]

where:
- \( P \) = pressure,
- \( M \) = molar mass,
- \( R \) = universal gas constant (0.0821 L·atm/(K·mol)),
- \( T \) = temperature in Kelvin.

### Step-by-step Solution:

1. **Identify the conditions for the gaseous oxide and dioxygen:**
- For the gaseous oxide:
- Pressure \( P_1 = 2 \) atm
- Temperature \( T = 273 \) K
- For dioxygen (O₂):
- Pressure \( P_2 = 5 \) atm
- Temperature \( T = 273 \) K

2. **Set up the density equations for both gases:**
- Density of the oxide:
\[ d_{oxide} = \frac{P_1 M_{oxide}}{RT} \]
- Density of dioxygen:
\[ d_{O_2} = \frac{P_2 M_{O_2}}{RT} \]

3. **Since the densities are equal, set the two equations equal to each other:**
\[ \frac{P_1 M_{oxide}}{RT} = \frac{P_2 M_{O_2}}{RT} \]

4. **Cancel out \( R \) and \( T \) from both sides:**
\[ P_1 M_{oxide} = P_2 M_{O_2} \]

5. **Substitute the known values:**
- The molar mass of dioxygen \( M_{O_2} = 32 \) g/mol (since O has a molar mass of 16 g/mol and O₂ has two oxygen atoms).
\[ 2 \, \text{atm} \cdot M_{oxide} = 5 \, \text{atm} \cdot 32 \, \text{g/mol} \]

6. **Solve for \( M_{oxide} \):**
\[ M_{oxide} = \frac{5 \cdot 32}{2} \]
\[ M_{oxide} = \frac{160}{2} \]
\[ M_{oxide} = 80 \, \text{g/mol} \]

### Final Answer:
The molecular mass of the oxide is **80 g/mol**.