Equal weights of `SO_2` and `SO_3` are present in a flask. Mole fraction of `SO_3` are present in a flask. Mole fraction of `SO_2` in the mixture

Options

A. Is greater than that of `SO_3`
B. Is smaller than that of `SO_3`
C. Is equal to that of `SO_3`
D. Is smaller than that of `C_5H_12`, if `SO_3` is replaced by same amount of `C_5H_12`

Answer

To find the mole fraction of \( SO_2 \) in a mixture where equal weights of \( SO_2 \) and \( SO_3 \) are present, we can follow these steps:

### Step 1: Determine the Molecular Weights
- The molecular weight of \( SO_2 \):
- Sulfur (S) = 32 g/mol
- Oxygen (O) = 16 g/mol
- Therefore, \( SO_2 = 32 + (2 \times 16) = 32 + 32 = 64 \) g/mol

- The molecular weight of \( SO_3 \):
- Sulfur (S) = 32 g/mol
- Oxygen (O) = 16 g/mol
- Therefore, \( SO_3 = 32 + (3 \times 16) = 32 + 48 = 80 \) g/mol

### Step 2: Calculate the Number of Moles
- Let the weight of \( SO_2 \) and \( SO_3 \) be \( W \) grams each.
- The number of moles of \( SO_2 \):
\[ n_{SO_2} = \frac{W}{64} \]
- The number of moles of \( SO_3 \):
\[ n_{SO_3} = \frac{W}{80} \]

### Step 3: Calculate the Total Number of Moles
- The total number of moles in the mixture:
\[ n_{total} = n_{SO_2} + n_{SO_3} = \frac{W}{64} + \frac{W}{80} \]
- To add these fractions, we need a common denominator. The least common multiple of 64 and 80 is 320.
\[ n_{total} = \frac{5W}{320} + \frac{4W}{320} = \frac{(5W + 4W)}{320} = \frac{9W}{320} \]

### Step 4: Calculate the Mole Fraction of \( SO_2 \)
- The mole fraction of \( SO_2 \) is given by:
\[ X_{SO_2} = \frac{n_{SO_2}}{n_{total}} = \frac{\frac{W}{64}}{\frac{9W}{320}} \]
- Simplifying this expression:
\[ X_{SO_2} = \frac{W}{64} \times \frac{320}{9W} = \frac{320}{576} = \frac{5}{9} \]

### Final Answer
The mole fraction of \( SO_2 \) in the mixture is \( \frac{5}{9} \).

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