At `0^(@)C` the density of a gaseous oxide at 2 bar is same as that of nitrogen at 5 bar What is the molecular mass of the oxide ? .

Explanation

To find the molecular mass of the gaseous oxide, we can use the relationship between density, pressure, molecular mass, and temperature. Here’s a step-by-step solution:

### Step 1: Understand the given information
We know that:
- The density of the gaseous oxide at 2 bar is equal to the density of nitrogen at 5 bar.
- The temperature for both gases is 0°C (which is 273.15 K).
- The molecular mass of nitrogen (M2) is 28 g/mol.

### Step 2: Write the formula for density
The formula for density (D) of a gas is given by:
\[ D = \frac{PM}{RT} \]
where:
- \( P \) = pressure,
- \( M \) = molecular mass,
- \( R \) = universal gas constant (approximately 0.0831 L·bar/K·mol),
- \( T \) = temperature in Kelvin.

### Step 3: Set up the equation for the densities
Since the densities of the gaseous oxide and nitrogen are equal, we can write:
\[ \frac{P_1 M_1}{RT_1} = \frac{P_2 M_2}{RT_2} \]

### Step 4: Simplify the equation
Since the temperature (T) and the gas constant (R) are the same for both gases, they cancel out:
\[ P_1 M_1 = P_2 M_2 \]

### Step 5: Substitute the known values
Let:
- \( P_1 = 2 \) bar (pressure of the gaseous oxide),
- \( M_1 \) = molecular mass of the gaseous oxide (what we want to find),
- \( P_2 = 5 \) bar (pressure of nitrogen),
- \( M_2 = 28 \) g/mol (molecular mass of nitrogen).

So, we have:
\[ 2 M_1 = 5 \times 28 \]

### Step 6: Calculate \( M_1 \)
Now, calculate \( M_1 \):
\[ 2 M_1 = 140 \]
\[ M_1 = \frac{140}{2} \]
\[ M_1 = 70 \text{ g/mol} \]

### Final Answer
The molecular mass of the gaseous oxide is **70 g/mol**.

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