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 `35^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 equation rearranged for molar mass. 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: 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 for molar mass \( M \)
Rearranging the equation to solve for molar mass \( M \):
\[ M = \frac{mRT}{PV} \]

### Step 5: Use density to find mass
Since density \( d \) is defined as:
\[ d = \frac{m}{V} \]
We can express mass \( m \) as:
\[ m = d \cdot V \]
Substituting this into the molar mass equation gives:
\[ M = \frac{d \cdot V \cdot RT}{PV} \]

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

### Step 7: Substitute the known values
Now we can substitute the known values into the equation:
- Density \( d = 3.380 \, \text{g/L} \)
- Temperature \( T = 35^\circ C = 308 \, \text{K} \) (convert Celsius to Kelvin by adding 273)
- Pressure \( P = 1 \, \text{atm} \)
- Gas constant \( R = 0.0821 \, \text{L·atm/(K·mol)} \)

Substituting these values:
\[ M = \frac{3.380 \, \text{g/L} \times 0.0821 \, \text{L·atm/(K·mol)} \times 308 \, \text{K}}{1 \, \text{atm}} \]

### Step 8: Calculate the molar mass
Now, calculating the value:
\[ M = \frac{3.380 \times 0.0821 \times 308}{1} \]
\[ M = \frac{85.46}{1} \]
\[ M \approx 85.46 \, \text{g/mol} \]

### Final Answer
The molar mass of the gas is approximately **85.46 g/mol**.

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