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 3.380 `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 using the given density, temperature, and pressure, we can use the ideal gas law in conjunction with the formula for density. Here’s a 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
- \( R \) = ideal gas constant (0.0821 L·atm/(K·mol))
- \( T \) = temperature (in Kelvin)

### Step 2: Express the number of moles (n)
The number of moles 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 n in the Ideal Gas Law
Substituting the expression for \( n \) into the ideal gas law gives:
\[ PV = \frac{m}{M}RT \]

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

### Step 5: Relate mass to density
Since density \( d \) is defined as mass per unit volume:
\[ d = \frac{m}{V} \]
Thus, we can express mass as:
\[ m = dV \]

### Step 6: Substitute mass into the equation for M
Substituting \( m \) into the equation for \( M \):
\[ M = \frac{dVRT}{PV} \]

### Step 7: Cancel out volume (V)
The volume \( V \) cancels out:
\[ M = \frac{dRT}{P} \]

### Step 8: Substitute known values
Now, substitute the known values into the equation:
- Density \( d = 3.380 \, g/L \)
- \( R = 0.0821 \, L \cdot atm/(K \cdot mol) \)
- Temperature \( T = 15^\circ C = 15 + 273 = 288 \, K \)
- Pressure \( P = 1 \, atm \)

Substituting these values gives:
\[ M = \frac{3.380 \, g/L \times 0.0821 \, L \cdot atm/(K \cdot mol) \times 288 \, K}{1 \, atm} \]

### Step 9: Calculate the Molar Mass
Calculating the above expression:
\[ M = \frac{3.380 \times 0.0821 \times 288}{1} \]
\[ M = \frac{79.9 \, g \cdot mol^{-1}}{1} \]
Thus, the molar mass of the gas is approximately:
\[ M \approx 79.9 \, g/mol \]

### Final Answer
The molar mass of the gas is approximately **79.9 g/mol**.

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