A gas cylinder was found unattended in a public place. The investigating team took the collected samples from it. The density of the gas was found to be 2.380 `gL^(−1)` at `35^o`C and 1 atm pressure. Hence the molar mass of the gas is:

Options

A. 20g/mol
B. 40g/mol
C. 60g/mol
D. 50g/mol

Explanation

To find the molar mass of the gas using the given density, temperature, and pressure, we can apply 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. **Convert the Temperature to Kelvin:**
\[ T = 35^\circ C + 273.15 = 308.15 \, K \]

2. **Use the Ideal Gas Law to Relate Density to Molar Mass:**
The number of moles \( n \) can be expressed as:
\[ n = \frac{\text{mass}}{\text{molar mass}} = \frac{m}{M} \]
Where:
- \( m \) = mass of the gas (in grams)
- \( M \) = molar mass (in g/mol)

Substituting \( n \) into the ideal gas law gives:
\[ PV = \frac{m}{M}RT \]

3. **Rearranging the Equation:**
Rearranging the equation to solve for molar mass \( M \):
\[ M = \frac{mRT}{PV} \]

4. **Using Density to Find Mass:**
The density \( d \) of the gas is given as \( 2.380 \, g/L \). Therefore, we can express mass \( m \) in terms of density and volume \( V \):
\[ m = d \cdot V \]

5. **Substituting Density into the Molar Mass Equation:**
Substituting \( m \) into the molar mass equation:
\[ M = \frac{d \cdot V \cdot RT}{PV} \]
The volume \( V \) cancels out:
\[ M = \frac{dRT}{P} \]

6. **Plugging in the Values:**
Now, substitute the known values:
- Density \( d = 2.380 \, g/L \)
- Pressure \( P = 1 \, atm \)
- Gas constant \( R = 0.0821 \, L \cdot atm/(K \cdot mol) \)
- Temperature \( T = 308.15 \, K \)

Thus, we have:
\[ M = \frac{2.380 \, g/L \cdot 0.0821 \, L \cdot atm/(K \cdot mol) \cdot 308.15 \, K}{1 \, atm} \]

7. **Calculating Molar Mass:**
\[ M = \frac{2.380 \cdot 0.0821 \cdot 308.15}{1} \]
\[ M \approx \frac{60.00}{1} \approx 60 \, g/mol \]

### Final Answer:
The molar mass of the gas is approximately **60 g/mol**.

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