Calculate the total pressure in a mixture of $8 \,g$ of dioxygen and $4 \,g$ of dihydrogen confined in a vessel of $1 \,dm^{3}$ at $27^{\circ} C$. $(R = 0.083 \,bar \,dm^{3} \,K^{-1} \,mol^{-1})$

(N/A) Given,
Mass of dioxygen $(O_{2}) = 8 \,g$
Number of moles of $O_{2} = \frac{8 \,g}{32 \,g \,mol^{-1}} = 0.25 \,mol$
Mass of dihydrogen $(H_{2}) = 4 \,g$
Number of moles of $H_{2} = \frac{4 \,g}{2 \,g \,mol^{-1}} = 2 \,mol$
Total number of moles $(n) = 0.25 \,mol + 2 \,mol = 2.25 \,mol$
Volume $(V) = 1 \,dm^{3}$
Temperature $(T) = 27 + 273 = 300 \,K$
Gas constant $(R) = 0.083 \,bar \,dm^{3} \,K^{-1} \,mol^{-1}$
Using the ideal gas equation,$pV = nRT$:
$p = \frac{nRT}{V} = \frac{2.25 \,mol \times 0.083 \,bar \,dm^{3} \,K^{-1} \,mol^{-1} \times 300 \,K}{1 \,dm^{3}}$
$p = 56.025 \,bar$
The total pressure of the mixture is $56.025 \,bar$.