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 1.380 `gL^(−1)` at `25^o`C and 1 atm pressure. Hence the molar mass of the gas is:

Options

A. 33.72g/mol
B. 14.72g/mol
C. 28g/mol
D. 25g/mol

Explanation

To find the molar mass of the gas using the given density, we can use the ideal gas equation, 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 atm L K\(^{-1}\) mol\(^{-1}\))
- \( T \) = Temperature (in Kelvin)

### Step-by-Step Solution:

1. **Convert Temperature to Kelvin:**
\[ T = 25^\circ C + 273.15 = 298.15 \, K \]

2. **Use the Ideal Gas Equation to Relate Density and Molar Mass:**
The number of moles \( n \) can also be expressed as:
\[ n = \frac{mass}{molar \, mass} = \frac{W}{M} \]
Where \( W \) is the mass of the gas and \( M \) is the molar mass.

Rearranging the ideal gas equation gives:
\[ PV = \frac{W}{M}RT \]
Rearranging for molar mass \( M \):
\[ M = \frac{WRT}{PV} \]

3. **Express Mass in Terms of Density:**
The mass \( W \) can be expressed as:
\[ W = \text{Density} \times V \]
Therefore, substituting this into the equation for molar mass gives:
\[ M = \frac{(\text{Density} \times V)RT}{PV} \]
The volume \( V \) cancels out:
\[ M = \frac{\text{Density} \times RT}{P} \]

4. **Substitute the Known Values:**
Given:
- Density = 1.380 g/L
- \( R = 0.0821 \, \text{atm L K}^{-1} \text{mol}^{-1} \)
- \( T = 298.15 \, K \)
- \( P = 1 \, atm \)

Substituting these values into the equation:
\[ M = \frac{1.380 \, g/L \times 0.0821 \, atm \, L \, K^{-1} \, mol^{-1} \times 298.15 \, K}{1 \, atm} \]

5. **Calculate the Molar Mass:**
\[ M = \frac{1.380 \times 0.0821 \times 298.15}{1} \]
\[ M = 33.76 \, g/mol \]

### Final Answer:
The molar mass of the gas is approximately **33.76 g/mol**.

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