Given that the weight ratio (w/w) of $\text{H}_2$ to $\text{O}_2$ is $1:4$. Let the mass of $\text{H}_2$ present in the mixture be $w \text{ g}$. Then, the mass of $\text{O}_2$ present will be $4w \text{ g}$. The molar mass of $\text{H}_2$ is $2 \text{ g mol}^{-1}$ and the molar mass of $\text{O}_2$ is $32 \text{ g mol}^{-1}$ . Number of moles is calculated as the ratio of given mass to the molar mass . Number of moles of $\text{H}_2$ ($n_{\text{H}_2}$) $= \frac{\text{Mass of H}_2}{\text{Molar mass of H}_2} = \frac{w}{2}$ Number of moles of $\text{O}_2$ ($n_{\text{O}_2}$) $= \frac{\text{Mass of O}_2}{\text{Molar mass of O}_2} = \frac{4w}{32} = \frac{w}{8}$ Now, let us find the molar ratio of $\text{H}_2$ to $\text{O}_2$: $\text{Molar ratio} = \frac{n_{\text{H}_2}}{n_{\text{O}_2}} = \frac{\frac{w}{2}}{\frac{w}{8}} = \frac{w}{2} \times \frac{8}{w} = \frac{4}{1} = 4:1$. Therefore, the molar ratio of the two gases in the mixture is $4:1$.