A and `B` are two idential vessels. `A` contains `15 g` ethane at `1 atm` and `298 K`. The vessel `B` contains `75 g` of a gas `X_(2)` at same temperature and pressure. The vapour density of `X_(2)` is :

Options

A. `75`
B. `150`
C. `37.5`
D. `45`

Explanation

To find the vapor density of gas \(X_2\) in vessel \(B\), we can follow these steps:

### Step 1: Understand the Ideal Gas Law
The ideal gas law is given by the equation:
\[ PV = nRT \]
Where:
- \(P\) = pressure
- \(V\) = volume
- \(n\) = number of moles
- \(R\) = universal gas constant
- \(T\) = temperature

### Step 2: Set Up the Problem
Given that both vessels \(A\) and \(B\) are identical, and they are at the same pressure and temperature, we can conclude that:
- \(P_A = P_B\)
- \(T_A = T_B\)
- \(V_A = V_B\)

### Step 3: Calculate Moles of Ethane in Vessel A
We know that vessel \(A\) contains \(15 \, g\) of ethane. The molecular weight of ethane \((C_2H_6)\) is \(30 \, g/mol\).

Using the formula for moles:
\[ n_A = \frac{\text{mass}}{\text{molar mass}} = \frac{15 \, g}{30 \, g/mol} = 0.5 \, mol \]

### Step 4: Relate Moles of Gas \(X_2\) in Vessel B
Since the conditions are the same for both vessels, the number of moles of gas \(X_2\) in vessel \(B\) is also \(n_B = n_A = 0.5 \, mol\).

### Step 5: Calculate the Molar Mass of Gas \(X_2\)
We know that vessel \(B\) contains \(75 \, g\) of gas \(X_2\). Let the molar mass of \(X_2\) be \(M\).

Using the formula for moles:
\[ n_B = \frac{\text{mass}}{\text{molar mass}} \Rightarrow 0.5 \, mol = \frac{75 \, g}{M} \]

Rearranging gives:
\[ M = \frac{75 \, g}{0.5 \, mol} = 150 \, g/mol \]

### Step 6: Calculate the Vapor Density of Gas \(X_2\)
Vapor density is defined as:
\[ \text{Vapor Density} = \frac{\text{Molar Mass}}{2} \]
Substituting the molar mass of \(X_2\):
\[ \text{Vapor Density} = \frac{150 \, g/mol}{2} = 75 \, g/L \]

### Conclusion
The vapor density of gas \(X_2\) is \(75 \, g/L\).