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

Options

A. 32.6g
B. 45g
C. 42.6g
D. 24g

Explanation

To find the molar mass of the gas from the given density, we can use the Ideal Gas Law, which is expressed as:

\[ PV = nRT \]

Where:
- \( P \) = pressure (in atm)
- \( V \) = volume (in liters)
- \( n \) = number of moles
- \( R \) = ideal gas constant
- \( T \) = temperature (in Kelvin)

### Step-by-Step Solution:

1. **Convert the temperature from Celsius to Kelvin:**
\[ T(K) = T(°C) + 273.15 \]
Given \( T = 15°C \):
\[ T = 15 + 273.15 = 288.15 \, K \]

2. **Identify the given values:**
- Density of the gas \( d = 1.380 \, g/L \)
- Pressure \( P = 1 \, atm \)
- Gas constant \( R = 0.0821 \, atm \cdot L \cdot K^{-1} \cdot mol^{-1} \)

3. **Use the relationship between density and molar mass:**
The number of moles \( n \) can be expressed as:
\[ n = \frac{mass}{molar \, mass} \]
Therefore, we can rearrange the Ideal Gas Law to find molar mass \( M \):
\[ PV = \frac{mass}{M} \cdot RT \]
Rearranging gives:
\[ M = \frac{mass \cdot R \cdot T}{P \cdot V} \]

4. **Substituting density into the equation:**
Since \( density = \frac{mass}{volume} \), we can express mass as:
\[ mass = density \cdot volume \]
Substituting this into the equation for molar mass:
\[ M = \frac{density \cdot volume \cdot R \cdot T}{P \cdot volume} \]
The volume cancels out:
\[ M = \frac{density \cdot R \cdot T}{P} \]

5. **Substituting the known values:**
\[ M = \frac{1.380 \, g/L \cdot 0.0821 \, atm \cdot L \cdot K^{-1} \cdot mol^{-1} \cdot 288.15 \, K}{1 \, atm} \]

6. **Calculating the molar mass:**
\[ M = \frac{1.380 \cdot 0.0821 \cdot 288.15}{1} \]
\[ M = \frac{32.6 \, g}{mol} \]

### Final Answer:
The molar mass of the gas is approximately **32.6 g/mol**.

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