The density of the vapour of a substance at 1 atm and 500 K is `0.36 kgm^(-3)`. If molar mass of gas is `18 g mol^(-1)` the molar volume of gas is `5 xx 10^(a) m^(3)//mol` What is the value of a ?

Explanation

To solve the problem, we need to find the value of \( a \) in the expression for the molar volume of the gas, given as \( 5 \times 10^{a} \, m^{3}/mol \). We will use the relationship between density, mass, and volume, along with the provided data.

### Step-by-Step Solution:

1. **Convert the Molar Mass**:
The molar mass of the gas is given as \( 18 \, g/mol \). To use consistent units, we convert this to kilograms:
\[ \text{Molar Mass} = 18 \, g/mol = 0.018 \, kg/mol \]

2. **Use the Density Formula**:
The density (\( \rho \)) of the gas is given by the formula:
\[ \rho = \frac{\text{mass}}{\text{volume}} \]
Rearranging this gives us:
\[ \text{mass} = \rho \times \text{volume} \]

3. **Substituting Values**:
We know the density (\( \rho = 0.36 \, kg/m^{3} \)) and we can express the volume in terms of the molar volume (\( V_m \)):
\[ \text{mass} = 0.36 \, kg/m^{3} \times V_m \]
Since the mass can also be expressed in terms of molar mass and number of moles (\( n \)):
\[ \text{mass} = n \times \text{Molar Mass} \]
For 1 mole, \( n = 1 \):
\[ \text{mass} = 0.018 \, kg \]

4. **Setting Up the Equation**:
Equating the two expressions for mass:
\[ 0.36 \, kg/m^{3} \times V_m = 0.018 \, kg \]

5. **Solving for Molar Volume**:
Rearranging gives:
\[ V_m = \frac{0.018 \, kg}{0.36 \, kg/m^{3}} = 0.05 \, m^{3} \]

6. **Expressing Molar Volume in Scientific Notation**:
We can express \( 0.05 \, m^{3} \) in scientific notation:
\[ 0.05 \, m^{3} = 5 \times 10^{-2} \, m^{3}/mol \]

7. **Identifying the Value of \( a \)**:
From the expression \( 5 \times 10^{a} \, m^{3}/mol \), we see that \( a = -2 \).

### Final Answer:
The value of \( a \) is \( -2 \).