Given:
- Pure \( \text{SO}_3 \) gas is heated to \( 600^\circ \text{C} \).
- It dissociates into \( \text{SO}_2 \) and \( \text{O}_2 \) gases up to 50%.
- We need to find the average molar mass (\( M_{\text{avg}} \)) of the final sample.
Chemical equation for dissociation:
\[ 2 \text{SO}_3 \leftrightarrow 2 \text{SO}_2 + \text{O}_2 \]
Initial moles:
- Assume initially we have 2 moles of \( \text{SO}_3 \).
Dissociation:
- 50% dissociation of \( \text{SO}_3 \) means half of the \( \text{SO}_3 \) dissociates.
- Therefore, 1 mole of \( \text{SO}_3 \) dissociates into 1 mole of \( \text{SO}_2 \) and 0.5 mole of \( \text{O}_2 \).
Moles after dissociation:
- Remaining \( \text{SO}_3 \): \( 2 - 1 = 1 \) mole
- Formed \( \text{SO}_2 \): \( 1 \) mole
- Formed \( \text{O}_2 \): \( 0.5 \) mole
Total moles of gas mixture:
\[ 1 \text{ mole (SO}_3\text{) + 1 mole (SO}_2\text{) + 0.5 mole (O}_2\text{) = 2.5 \text{ moles}} \]
Molar masses:
- Molar mass of \( \text{SO}_3 \): \( 80 \text{ g/mol} \)
- Molar mass of \( \text{SO}_2 \): \( 64 \text{ g/mol} \)
- Molar mass of \( \text{O}_2 \): \( 32 \text{ g/mol} \)
Total mass of gas mixture:
\[ \text{Mass of } \text{SO}_3 = 1 \text{ mole} \times 80 \text{ g/mol} = 80 \text{ g} \]
\[ \text{Mass of } \text{SO}_2 = 1 \text{ mole} \times 64 \text{ g/mol} = 64 \text{ g} \]
\[ \text{Mass of } \text{O}_2 = 0.5 \text{ mole} \times 32 \text{ g/mol} = 16 \text{ g} \]
\[ \text{Total mass} = 80 \text{ g} + 64 \text{ g} + 16 \text{ g} = 160 \text{ g} \]
Average molar mass (\( M_{\text{avg}} \)):
\[ M_{\text{avg}} = \frac{\text{Total mass}}{\text{Total moles}} = \frac{160 \text{ g}}{2.5 \text{ moles}} = 64 \text{ g/mol} \]
So, the value of the average molar mass (\( M_{\text{avg}} \)) is:
\[ M_{\text{avg}} = 64 \text{ g/mol} \]
To find \( \frac{M_{\text{avg}}}{100} \):
\[ \frac{M_{\text{avg}}}{100} = \frac{64 \text{ g/mol}}{100} = 0.64 \text{ g/mol} \]
Therefore, the value of \( \frac{M_{\text{avg}}}{100} \) is:
\[ \frac{M_{\text{avg}}}{100} = 0.64 \text{ g/mol} \]
