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 `15^o`C and 1 atm pressure. Hence the molar mass of the gas is:

Explanation

To find the molar mass of the gas from the given density, we can use the ideal gas law and the relationship between density and molar mass. Here’s the step-by-step solution:

### Step 1: Write down the ideal gas law
The ideal gas law is given by the equation:
\[ PV = nRT \]
where:
- \( P \) = pressure (in atm)
- \( V \) = volume (in liters)
- \( n \) = number of moles
- \( R \) = universal gas constant (0.0821 atm L K\(^{-1}\) mol\(^{-1}\))
- \( T \) = temperature (in Kelvin)

### Step 2: Relate moles to mass and molar mass
The number of moles (\( n \)) can also be expressed as:
\[ n = \frac{m}{M} \]
where:
- \( m \) = mass of the gas (in grams)
- \( M \) = molar mass of the gas (in g/mol)

### Step 3: Substitute into the ideal gas law
Substituting \( n \) into the ideal gas law gives:
\[ PV = \frac{m}{M} RT \]

### Step 4: Rearrange to find molar mass
Rearranging the equation to solve for molar mass \( M \):
\[ M = \frac{mRT}{PV} \]

### Step 5: Express mass per volume as density
Since density (\( D \)) is defined as:
\[ D = \frac{m}{V} \]
we can express mass as:
\[ m = DV \]
Substituting this into the equation for molar mass gives:
\[ M = \frac{D V RT}{PV} \]

### Step 6: Simplify the equation
The volume (\( V \)) cancels out:
\[ M = \frac{DRT}{P} \]

### Step 7: Substitute the known values
Now we can substitute the known values:
- Density (\( D \)) = 2.380 g/L
- Pressure (\( P \)) = 1 atm
- Gas constant (\( R \)) = 0.0821 atm L K\(^{-1}\) mol\(^{-1}\)
- Temperature (\( T \)) = 15°C = 15 + 273 = 288 K

Substituting these values into the equation:
\[ M = \frac{(2.380 \, \text{g/L}) \times (0.0821 \, \text{atm L K}^{-1} \text{mol}^{-1}) \times (288 \, \text{K})}{1 \, \text{atm}} \]

### Step 8: Calculate the molar mass
Now, performing the calculation:
\[ M = \frac{(2.380) \times (0.0821) \times (288)}{1} \]
\[ M = \frac{58.0}{1} \]
\[ M \approx 58.0 \, \text{g/mol} \]

### Final Answer
The molar mass of the gas is approximately **58 g/mol**.

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