Calculate the molar mass of an unknown gas using Graham's Law, given its diffusion rate and the diffusion rate of oxygen (32 g/mol).

$$\left(\frac{47.15}{60.40}\right)^2 = \frac{32}{M_1}$$ $$(0.780629)^2 \approx 0.60938 = \frac{32}{M_1}$$

$$M_1 = \frac{32}{0.60938}$$ $$M_1 \approx 52.51 \text{ g/mol}$$

The relative molecular mass of the unknown gas is approximately 52.51 g/mol.

If the question literally meant the relative molecular mass of oxygen, then it is simply $2 \times 16 = 32 \text{ g/mol}$. However, this would make the diffusion rates irrelevant.

2)
We will again use Graham's Law of Diffusion.

$$\frac{r_{H_2}}{r_{O_2}} = \sqrt{\frac{M_{O_2}}{M_{H_2}}}$$

Given:

• Volume of hydrogen (H$_2$) = $450 \text{ cm}^3$
• Time for hydrogen (H$_2$) = $25 \text{ sec}$
• Molecular mass of hydrogen ($M_{H_2}$) = $2 \times 1 = 2 \text{ g/mol}$
• Molecular mass of oxygen ($M_{O_2}$) = $2 \times 16 = 32 \text{ g/mol}$
• Volume of oxygen (O$_2$) = $450 \text{ cm}^3$ (same volume)
• Time for oxygen ($t_{O_2}$) = ?

$$\frac{18}{\frac{450}{t_{O_2}}} = \sqrt{\frac{32}{2}}$$ $$\frac{18 \cdot t_{O_2}}{450} = \sqrt{16}$$ $$\frac{18 \cdot t_{O_2}}{450} = 4$$

$$18 \cdot t_{O_2} = 4 \times 450$$ $$18 \cdot t_{O_2} = 1800$$ $$t_{O_2} = \frac{1800}{18}$$ $$t_{O_2} = 100 \text{ sec}$$

It would take 100 seconds for the same volume of oxygen to diffuse.

The final answers are:

1. The relative molecular mass of the unknown gas is $\boxed{\text{52.51 g/mol}}$.
2. The time taken for oxygen to diffuse is $\boxed{\text{100 seconds}}$.