A non-reacting gaseous mixture contains` SO_2` and `SO_3` in the mass ratio of `1 : 5`. Find the ratio of the number of molecules

Options

A. ` 1 : 1`
B. `4 : 5`
C. ` 1 : 4`
D. ` 1 : 5`

Answer

To solve the problem of finding the ratio of the number of molecules of \( SO_2 \) and \( SO_3 \) in a non-reacting gaseous mixture where the mass ratio is given as \( 1:5 \), we can follow these steps:

### Step 1: Define the Masses
Let the mass of \( SO_2 \) be \( m_{SO_2} = 1 \) g (as per the mass ratio) and the mass of \( SO_3 \) be \( m_{SO_3} = 5 \) g.

### Step 2: Calculate Molar Masses
The molar masses of the gases are:
- Molar mass of \( SO_2 \) (Sulfur Dioxide) = \( 32 \, \text{g/mol (S)} + 2 \times 16 \, \text{g/mol (O)} = 32 + 32 = 64 \, \text{g/mol} \)
- Molar mass of \( SO_3 \) (Sulfur Trioxide) = \( 32 \, \text{g/mol (S)} + 3 \times 16 \, \text{g/mol (O)} = 32 + 48 = 80 \, \text{g/mol} \)

### Step 3: Calculate the Number of Moles
Using the formula for the number of moles:
\[ \text{Number of moles} = \frac{\text{mass}}{\text{molar mass}} \]
- For \( SO_2 \):
\[ n_{SO_2} = \frac{m_{SO_2}}{M_{SO_2}} = \frac{1 \, \text{g}}{64 \, \text{g/mol}} = \frac{1}{64} \, \text{mol} \]
- For \( SO_3 \):
\[ n_{SO_3} = \frac{m_{SO_3}}{M_{SO_3}} = \frac{5 \, \text{g}}{80 \, \text{g/mol}} = \frac{5}{80} = \frac{1}{16} \, \text{mol} \]

### Step 4: Calculate the Ratio of Moles
Now, we find the ratio of the number of moles of \( SO_2 \) to \( SO_3 \):
\[ \text{Ratio of moles} = \frac{n_{SO_2}}{n_{SO_3}} = \frac{\frac{1}{64}}{\frac{1}{16}} = \frac{1}{64} \times \frac{16}{1} = \frac{16}{64} = \frac{1}{4} \]

### Step 5: Ratio of Number of Molecules
Since the number of molecules is directly proportional to the number of moles (Avogadro's law), the ratio of the number of molecules of \( SO_2 \) to \( SO_3 \) is the same as the ratio of the number of moles:
\[ \text{Ratio of molecules} = \frac{n_{SO_2}}{n_{SO_3}} = \frac{1}{4} \]

### Final Answer
The ratio of the number of molecules of \( SO_2 \) to \( SO_3 \) is \( 1:4 \).

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