`20g` of ideal gas contains only atoms of `S and O` occupies `5.6 L` at 1 atm and `273K.` what is the molecular mass of gas ?

Options

A. 64
B. 80
C. 96
D. None of these

Answer

To find the molecular mass of the ideal gas containing only sulfur (S) and oxygen (O), we can use the Ideal Gas Law, which is expressed as:

\[ PV = nRT \]

Where:
- \( P \) = pressure (in atm)
- \( V \) = volume (in liters)
- \( n \) = number of moles
- \( R \) = ideal gas constant (0.0821 L·atm/(K·mol))
- \( T \) = temperature (in Kelvin)

### Step-by-Step Solution:

1. **Identify Given Values:**
- Mass of gas, \( W = 20 \, \text{g} \)
- Volume, \( V = 5.6 \, \text{L} \)
- Pressure, \( P = 1 \, \text{atm} \)
- Temperature, \( T = 273 \, \text{K} \)

2. **Use the Ideal Gas Law:**
We will rearrange the Ideal Gas Law to solve for the number of moles \( n \):
\[ n = \frac{PV}{RT} \]

3. **Substitute the Known Values:**
Substitute the values into the equation:
\[ n = \frac{(1 \, \text{atm}) \times (5.6 \, \text{L})}{(0.0821 \, \text{L·atm/(K·mol)}) \times (273 \, \text{K})} \]

4. **Calculate the Number of Moles \( n \):**
\[ n = \frac{5.6}{0.0821 \times 273} \]
Calculate \( 0.0821 \times 273 \):
\[ 0.0821 \times 273 \approx 22.4143 \]
Now calculate \( n \):
\[ n = \frac{5.6}{22.4143} \approx 0.249 \, \text{moles} \]

5. **Calculate the Molecular Mass:**
The molecular mass \( M \) can be calculated using the formula:
\[ M = \frac{W}{n} \]
Substitute the known values:
\[ M = \frac{20 \, \text{g}}{0.249 \, \text{moles}} \approx 80.32 \, \text{g/mol} \]

6. **Final Result:**
The molecular mass of the gas is approximately \( 80 \, \text{g/mol} \).