The density of vapours of a substance at 1 atm and 500 K is `0.3 kg m^(-3)`. The vapours effuse 0.4216 times faster than `O_(2)` through a pin hole under identical conditions. If R= 0.08 litre atm `K^(-1) mol^(-1)`. The molar volume of gas is `a xx 10^(2)` litre. The value of a is _____________.

Explanation

To solve the problem step by step, we will follow the given information and apply the relevant concepts from the states of matter and gas laws.

### Step 1: Understand the relationship of effusion rates
The problem states that the vapors effuse 0.4216 times faster than \( O_2 \). According to Graham's law of effusion, the rate of effusion of a gas is inversely proportional to the square root of its molar mass. This can be expressed as:

\[ \frac{R_{vapor}}{R_{O_2}} = \sqrt{\frac{M_{O_2}}{M_{vapor}}} \]

Given that \( R_{vapor} = 0.4216 \times R_{O_2} \) and the molar mass of \( O_2 \) is 32 g/mol, we can set up the equation:

\[ 0.4216 = \sqrt{\frac{32}{M_{vapor}}} \]

### Step 2: Solve for the molar mass of the vapor
Squaring both sides gives:

\[ (0.4216)^2 = \frac{32}{M_{vapor}} \]

Calculating \( (0.4216)^2 \):

\[ 0.1774 = \frac{32}{M_{vapor}} \]

Now, rearranging to find \( M_{vapor} \):

\[ M_{vapor} = \frac{32}{0.1774} \approx 180.0 \text{ g/mol} \]

### Step 3: Convert the molar mass to kg/mol
Since we need the molar mass in kg/mol for further calculations:

\[ M_{vapor} = 180.0 \text{ g/mol} = 0.180 \text{ kg/mol} \]

### Step 4: Use the density to find the molar volume
The density of the vapor is given as \( 0.3 \text{ kg/m}^3 \). The molar volume \( V_m \) can be calculated using the formula:

\[ \text{Density} = \frac{\text{mass}}{\text{volume}} \Rightarrow V_m = \frac{M_{vapor}}{\text{Density}} \]

Substituting the values:

\[ V_m = \frac{0.180 \text{ kg/mol}}{0.3 \text{ kg/m}^3} \]

Calculating this gives:

\[ V_m = 0.6 \text{ m}^3/\text{mol} \]

### Step 5: Convert the molar volume to liters
Since \( 1 \text{ m}^3 = 1000 \text{ liters} \):

\[ V_m = 0.6 \text{ m}^3/\text{mol} \times 1000 \text{ L/m}^3 = 600 \text{ L/mol} \]

### Step 6: Express the molar volume in the required form
The problem states that the molar volume can be expressed as \( a \times 10^2 \text{ L} \). Thus:

\[ 600 \text{ L/mol} = 6 \times 10^2 \text{ L/mol} \]

### Conclusion
The value of \( a \) is:

\[ \boxed{6} \]