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

Options

A. 17g/mol
B. 34g/mol
C. 28g/mol
D. 26g/mol

Explanation

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

### Step 1: Understand 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 atm L K\(^{-1}\) mol\(^{-1}\))
- \( T \) = temperature (in Kelvin)

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

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

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

### Step 4: Express Mass in Terms of Density
The mass \( m \) can be expressed as:
\[ m = \text{density} \times V \]
Thus, substituting this into the equation gives:
\[ M = \frac{\text{density} \times V \times RT}{PV} \]

### Step 5: Simplifying the Equation
Notice that \( V \) cancels out:
\[ M = \frac{\text{density} \times RT}{P} \]

### Step 6: Plugging in the Values
Now, we can plug in the values:
- Density = 1.380 g/L
- \( R = 0.0821 \) atm L K\(^{-1}\) mol\(^{-1}\)
- \( P = 1 \) atm
- Temperature = \( 27^\circ C = 300 \) K (convert Celsius to Kelvin by adding 273)

So the equation becomes:
\[ M = \frac{1.380 \, \text{g/L} \times 0.0821 \, \text{atm L K}^{-1} \text{mol}^{-1} \times 300 \, \text{K}}{1 \, \text{atm}} \]

### Step 7: Calculating Molar Mass
Now, calculate:
\[ M = \frac{1.380 \times 0.0821 \times 300}{1} \]
\[ M = 33.98 \, \text{g/mol} \]

### Step 8: Rounding Off
Rounding off gives:
\[ M \approx 34 \, \text{g/mol} \]

### Conclusion
The molar mass of the gas is approximately **34 g/mol**.

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