The density of the vapour of a substance at `1 atm` pressure and `500 K` is `0.36 kg m^(-3)`. The vapour effuses through a small hole at a rate of `1.33` times faster than oxygen under the same condition.
(`a`) Determine (`i`) the molecular weight, (`ii`) the molar volume

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 diffusion.

### Step 1: Determine the Molecular Weight using Graham's Law of Effusion

1. **Understanding Graham's Law**: Graham's Law states that the rate of effusion of a gas is inversely proportional to the square root of its molar mass. Mathematically, it can be expressed as:
\[ \frac{R_1}{R_2} = \sqrt{\frac{M_2}{M_1}} \]
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