p>$$ P_i = X_i \cdot P_{\text{total}} $$

where $ P_i $ is the partial pressure of the gas, $ X_i $ is the mole fraction of the gas, and $ P_{\text{total}} $ is the total pressure of the mixture.

Step 1: Calculate moles of each gas

Moles of $\mathrm{O}_2$:

$$ \text{Molar mass of } \mathrm{O}_2 = 32 \, \text{g/mol} $$

$$ n_{\mathrm{O}_2} = \frac{32 \, \text{g}}{32 \, \text{g/mol}} = 1 \, \text{mol} $$

Moles of Ar (Argon):

$$ \text{Molar mass of Ar} = 40 \, \text{g/mol} $$

$$ n_{\mathrm{Ar}} = \frac{80 \, \text{g}}{40 \, \text{g/mol}} = 2 \, \text{mol} $$

Moles of $\mathrm{H}_2$ (Dihydrogen):

$$ \text{Molar mass of } \mathrm{H}_2 = 2 \, \text{g/mol} $$

$$ n_{\mathrm{H}_2} = \frac{4 \, \text{g}}{2 \, \text{g/mol}} = 2 \, \text{mol} $$

Step 2: Calculate total moles of gas mixture

$$ n_{\text{total}} = n_{\mathrm{O}_2} + n_{\mathrm{Ar}} + n_{\mathrm{H}_2} = 1 \, \text{mol} + 2 \, \text{mol} + 2 \, \text{mol} = 5 \, \text{mol} $$

Step 3: Calculate mole fraction of $\mathrm{O}_2$

$$ X_{\mathrm{O}_2} = \frac{n_{\mathrm{O}_2}}{n_{\text{total}}} = \frac{1 \, \text{mol}}{5 \, \text{mol}} = 0.2 $$

Step 4: Calculate partial pressure of $\mathrm{O}_2$

Given $ P_{\text{total}} = 10 \, \text{bar} $,

$$ P_{\mathrm{O}_2} = X_{\mathrm{O}_2} \times P_{\text{total}} = 0.2 \times 10 \, \text{bar} = 2 \, \text{bar} $$

Therefore, the partial pressure exerted by dioxygen is 2 bar.

Option B: 2 bar

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