To find the total pressure in the mixture, we will use the ideal gas law.$\\$Given:$\\$• Mass of $O_2 = 4$ g$\\$• Mass of $H_2 = 2$ g$\\$• Total volume $V = 1$ L$\\$• Temperature $T = 273$ K$\\$• Gas constant $R = 0.082$ L atm mol$^{-1}$ K$^{-1} \\$First, we need to calculate the number of moles of each gas.$\\$For $O_2: \\$Molar mass of $O_2 = 32$ g/mol$\\ n_{O_2} = \frac{4}{32} = 0.125$ mol$\\$For $H_2: \\$Molar mass of $H_2 = 2$ g/mol$\\ n_{H_2} = \frac{2}{2} = 1$ mol$\\$Total number of moles $n_{total} = n_{O_2} + n_{H_2} = 0.125 + 1 = 1.125$ mol$\\$Using the ideal gas equation $PV = nRT,$ we can find the total pressure $P: \\ P = \frac{n_{total} \cdot R \cdot T}{V} \\ P = \frac{1.125 \cdot 0.082 \cdot 273}{1} \\ P = 25.18$ atm$\\$Therefore, the total pressure in the mixture is $25.18$ atm.$\\$This corresponds to Option 3.